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Solve exponential equations

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Difficulty: 1 – 5

51009612

Which expression gives the solution of \(2^{6x}=\frac{1}{3}\)? a) \(x=6\log_2\left(\frac{1}{3}\right)\) b) \(x=\frac{1}{6}\log_{1/3}(2)\) c) \(x=\frac{1}{3}\div2^6\) d) \(x=-\frac{1}{6}\log_2(3)\)

Hints

- Use a logarithm to bring the variable out of the exponent. - A logarithm with base \(2\) matches the exponential base. - Use the quotient property for the logarithm of a reciprocal.

Solution

1. Take the logarithm with base \(2\) of both sides: \(\log_2(2^{6x})=\log_2\left(\frac{1}{3}\right)\). 2. Simplify and solve: \(6x=\log_2\left(\frac{1}{3}\right)\), so \(x=\frac{1}{6}\log_2\left(\frac{1}{3}\right)\). 3. Since \(\log_2\left(\frac{1}{3}\right)=-\log_2(3)\), \(x=-\frac{1}{6}\log_2(3)\). 4. Therefore, choice d is correct.

Answer

d) \(x=-\frac{1}{6}\log_2(3)\)

52332312

Solve each exponential equation. a) \(5^{x-3}=125\) b) \(4\cdot10^x+50=40{,}050\) c) \(2^{3x+1}=128\)

Hints

- Rewrite both sides as powers with the same base when possible. - Isolate the exponential expression first. - Equal powers with the same valid base have equal exponents. - Identify the exponent that produces the number on the other side.

Solution

1. a) Rewrite \(125\) as \(5^3\). Then \(5^{x-3}=5^3\), so \(x-3=3\) and \(x=6\). 2. b) Isolate the exponential term: \(4\cdot10^x=40{,}000\), so \(10^x=10{,}000=10^4\). Therefore, \(x=4\). 3. c) Rewrite \(128\) as \(2^7\). Then \(2^{3x+1}=2^7\), so \(3x+1=7\) and \(x=2\).

Answer

a) \(x=6\) b) \(x=4\) c) \(x=2\)

52379112

Solve each exponential equation for \(x\). a) \(2\cdot10^x+50=250\) b) \(7^{x-2}=\frac{1}{49}\)

Hints

- Rewrite both sides with the same base when possible. - Express a reciprocal as a power with a negative exponent. - Isolate the exponential expression before comparing exponents.

Solution

1. a) Isolate the exponential term: \(2\cdot10^x=200\), so \(10^x=100=10^2\). Therefore, \(x=2\). 2. b) Since \(49=7^2\), \(\frac{1}{49}=7^{-2}\). Thus, \(7^{x-2}=7^{-2}\), so \(x-2=-2\) and \(x=0\).

Answer

a) \(x=2\) b) \(x=0\)

52605312

Let \(f(x)=e^x\). a) Evaluate the function at \(x_1=-1.2\), \(x_2=\sqrt{3}\), and \(x_3=\frac{2}{3}\). Round each value to four decimal places. b) Find the exact value of \(x\) for which \(f(x)\) has each given value: (1) \(y=\frac{e}{\sqrt[3]{e}}\) (2) \(y=\frac{1}{e^5}\) (3) \(y=e^2\sqrt{e}\)

Hints

- Can you rewrite each given value as a power with base \(e\)? - Which exponent rules help rewrite quotients and radicals? - What must be true of the exponents when two powers with the same positive base are equal? - Check the fifth decimal place before rounding to four decimal places.

Solution

1. Using a calculator, \(f(-1.2)=e^{-1.2}\approx0.3012\), \(f(\sqrt{3})=e^{\sqrt{3}}\approx5.6522\), and \(f\!\left(\frac{2}{3}\right)=e^{2/3}\approx1.9477\). 2. Rewrite each target value as a single power of \(e\): \(\frac{e}{\sqrt[3]{e}}=e^{1-1/3}=e^{2/3}\), \(\frac{1}{e^5}=e^{-5}\), and \(e^2\sqrt{e}=e^{2+1/2}=e^{5/2}\). 3. Since equal powers with base \(e\) have equal exponents, the corresponding values are \(x=\frac{2}{3}\), \(x=-5\), and \(x=\frac{5}{2}\).

Answer

a) \(f(-1.2)\approx0.3012\); \(f(\sqrt{3})\approx5.6522\); \(f\!\left(\frac{2}{3}\right)\approx1.9477\) b) (1) \(x=\frac{2}{3}\) (2) \(x=-5\) (3) \(x=\frac{5}{2}\)

52605412

Let \(g(x)=e^{x+1}\). a) Find the exact function values at \(x=-1\) and \(x=\ln(2)-1\). Also find the value at \(x=0\), rounded to four decimal places. b) Points on the graph of \(g\) have the following y-coordinates. Find the corresponding x-coordinate in each case: (1) \(y=e^4\) (2) \(y=\frac{1}{e}\) (3) \(y=\sqrt{e}\)

Hints

- Simplify the exponent after substituting each given x-value. - Recall the inverse relationship between \(e^x\) and \(\ln(x)\). - Rewrite each y-value as a power of \(e\). - Express a radical or reciprocal using a rational or negative exponent.

Solution

1. Substitute each input: \(g(-1)=e^{-1+1}=e^0=1\), \(g(\ln(2)-1)=e^{\ln(2)}=2\), and \(g(0)=e\approx2.7183\). 2. Write the target values as powers of \(e\): \(e^{x+1}=e^4\), \(e^{x+1}=e^{-1}\), and \(e^{x+1}=e^{1/2}\). 3. Equate the exponents. This gives \(x+1=4\), \(x+1=-1\), and \(x+1=\frac{1}{2}\), so \(x=3\), \(x=-2\), and \(x=-\frac{1}{2}\), respectively.

Answer

a) \(g(-1)=1\); \(g(\ln(2)-1)=2\); \(g(0)\approx2.7183\) b) (1) \(x=3\) (2) \(x=-2\) (3) \(x=-\frac{1}{2}\)

52610112

Let \(f(x)=e^x+4\) and \(g(x)=3-e^x\). Determine algebraically whether the graphs of \(f\) and \(g\) have any points in common.

Hints

- What equation represents an intersection of two function graphs? - Isolate the exponential expression. - Recall the range of \(e^x\). Can it equal a negative number?

Solution

1. At an intersection, the function values are equal: \(e^x+4=3-e^x\). 2. Combine the exponential terms to obtain \(2e^x=-1\). 3. Dividing by \(2\) gives \(e^x=-0.5\). 4. Since \(e^x>0\) for every real \(x\), this equation has no real solution. 5. Therefore, the graphs have no points in common.

Answer

The graphs have no points in common.

52625112

Solve each equation over the real numbers. a) \(e^{3x+5}=\frac{1}{e}\) b) \(e^{2-x}=\sqrt[3]{e^2}\) c) \(e^{x^2-9}=1\)

Hints

- Rewrite each right-hand side as a power of \(e\). - Equal powers with the same positive base have equal exponents. - Remember that the quadratic equation in part c has two solutions.

Solution

1. a) Write \(\frac{1}{e}=e^{-1}\). Then \(3x+5=-1\), so \(x=-2\). 2. b) Write \(\sqrt[3]{e^2}=e^{2/3}\). Then \(2-x=\frac{2}{3}\), so \(x=\frac{4}{3}\). 3. c) Write \(1=e^0\). Then \(x^2-9=0\), so \(x=-3\) or \(x=3\).

Answer

a) \(\{-2\}\) b) \(\left\{\frac{4}{3}\right\}\) c) \(\{-3,3\}\)

52626512

Solve \(5e^{0.5x}-12=8\). Give an exact answer using the natural logarithm and an approximation rounded to three decimal places.

Hints

- Isolate the exponential expression. - Apply the natural logarithm. - Solve the resulting linear equation.

Solution

1. Isolate the exponential term: \(5e^{0.5x}=20\), so \(e^{0.5x}=4\). 2. Take natural logarithms: \(0.5x=\ln(4)\). 3. Therefore, \(x=2\ln(4)=\ln(16)\approx2.773\).

Answer

\(x=2\ln(4)=\ln(16)\approx2.773\)

52644312

A bacterial culture grows exponentially. At the start of an observation period (\(t=0\)), the culture contains \(1200\) bacteria. After \(5\) hours, it contains \(3000\) bacteria. Find a function of the form \(f(t)=be^{kt}\) that models the number of bacteria after \(t\) hours.

Hints

- What does the value at \(t=0\) tell you about \(b\)? - Substitute the value at \(t=5\) into the model. - Use the natural logarithm to solve for the unknown in the exponent.

Solution

1. Since \(f(0)=1200\), the initial value is \(b=1200\). 2. Using \(f(5)=3000\) gives \(1200e^{5k}=3000\), so \(e^{5k}=2.5\). 3. Taking the natural logarithm gives \(5k=\ln(2.5)\), so \(k=\frac{\ln(2.5)}{5}\approx 0.1833\). 4. Therefore, the model is \(f(t)=1200e^{\frac{\ln(2.5)}{5}t}\), or approximately \(f(t)=1200e^{0.1833t}\).

Answer

\(f(t)=1200e^{\frac{\ln(2.5)}{5}t}\approx 1200e^{0.1833t}\)

52644412

An industrial machine loses \(12\%\) of its value each year. Its purchase price at \(t=0\) is \(\$45{,}000\). Write a function of the form \(f(t)=be^{kt}\) that models the machine's value after \(t\) years.

Hints

- Convert the annual percent decrease to a yearly multiplication factor. - Use \(e^k=a\) to relate the yearly factor to \(k\). - A decreasing model should have a negative value of \(k\).

Solution

1. The initial value is \(b=45{,}000\). 2. A yearly decrease of \(12\%\) gives a yearly factor of \(1-0.12=0.88\). 3. In the form \(e^{kt}\), \(e^k=0.88\), so \(k=\ln(0.88)\approx -0.1278\). 4. Therefore, \(f(t)=45{,}000e^{\ln(0.88)t}\approx 45{,}000e^{-0.1278t}\).

Answer

\(f(t)=45{,}000e^{\ln(0.88)t}\approx 45{,}000e^{-0.1278t}\)

52645712

A yeast culture covers an area of \(15\,\text{cm}^2\) at the start of an experiment (\(t=0\)). After \(6\) hours, it covers \(45\,\text{cm}^2\). The area is modeled by \(A(t)=A_0e^{kt}\), where \(t\) is measured in hours. 1. Find \(A_0\) and the exact value of \(k\). 2. Find the area covered after \(12\) hours.

Hints

- The value at \(t=0\) determines \(A_0\). - Substitute the second data point and use a logarithm to solve for \(k\). - Use logarithm and exponent properties to simplify the value at \(t=12\).

Solution

1. Since \(A(0)=A_0\), \(A_0=15\). 2. Using \(A(6)=45\) gives \(15e^{6k}=45\), so \(e^{6k}=3\). Taking the natural logarithm gives \(6k=\ln 3\), so \(k=\frac{\ln 3}{6}\). 3. Then \(A(12)=15e^{(\ln 3/6)(12)}=15e^{2\ln 3}=15(3^2)=135\).

Answer

1. \(A_0=15\); \(k=\frac{\ln 3}{6}\) 2. \(135\,\text{cm}^2\)

52828112

Solve \(25^{x+1}=5\cdot125^x\) without using logarithms.

Hints

- Rewrite all numbers as powers with the same base. - Use \(5\) as the common base for \(5\), \(25\), and \(125\). - Apply exponent properties until each side is a single power. - When the bases match, set the exponents equal.

Solution

1. Rewrite every number as a power of \(5\): \((5^2)^{x+1}=5\cdot(5^3)^x\). 2. Apply exponent properties: \(5^{2x+2}=5^{3x+1}\). 3. Set the exponents equal: \(2x+2=3x+1\). 4. Solve the linear equation to obtain \(x=1\).

Answer

\(x=1\)

52840512

Solve for \(x\). \(12\cdot1.5^x=60.75\)

Hints

- Isolate the exponential expression first. - Apply a logarithm to bring the variable out of the exponent. - Use the power property of logarithms. - Substitute the result into the original equation to check it.

Solution

1. Divide by \(12\): \(1.5^x=5.0625\). 2. Apply a logarithm to both sides: \(x\log(1.5)=\log(5.0625)\). 3. Solve: \(x=\frac{\log(5.0625)}{\log(1.5)}=4\).

Answer

\(x=4\)

52841112

Solve for \(x\). \(3\cdot2.5^x-18.5=28.375\)

Hints

- Isolate the exponential expression with base \(2.5\). - Apply a logarithm after simplifying the equation. - Use change of base if your calculator does not have a base-\(2.5\) logarithm key.

Solution

1. Add \(18.5\): \(3\cdot2.5^x=46.875\). 2. Divide by \(3\): \(2.5^x=15.625\). 3. Apply a logarithm: \(x=\log_{2.5}(15.625)=\frac{\log(15.625)}{\log(2.5)}\). 4. Since \(2.5^3=15.625\), \(x=3\).

Answer

\(x=3\)

52841212

Solve the exponential equation for \(x\). \(12\cdot0.5^{x-2}+7=10\)

Hints

- Isolate the power with base \(0.5\). - Identify the exponent that makes \(0.5\) equal to \(0.25\). - Remember that the entire expression \(x-2\) is the exponent.

Solution

1. Subtract \(7\): \(12\cdot0.5^{x-2}=3\). 2. Divide by \(12\): \(0.5^{x-2}=0.25\). 3. Since \(0.25=0.5^2\), set the exponents equal: \(x-2=2\). 4. Therefore, \(x=4\).

Answer

\(x=4\)

52841912

Solve each exponential equation by rewriting both sides with a common base. a) \(3^{2x+5}=\frac{1}{27}\) b) \(4^{x-1}=0.125\)

Hints

- Rewrite each number as a power with a common base. - Use negative exponents for reciprocals and terminating decimals. - When the bases match, set the exponents equal. - Apply the power-of-a-power property to simplify the left side.

Solution

1. a) Rewrite \(\frac{1}{27}\) as \(3^{-3}\). Then \(2x+5=-3\), so \(2x=-8\) and \(x=-4\). 2. b) Rewrite both sides with base \(2\): \(4^{x-1}=2^{2x-2}\) and \(0.125=\frac{1}{8}=2^{-3}\). 3. Set the exponents equal: \(2x-2=-3\), so \(2x=-1\) and \(x=-\frac{1}{2}\).

Answer

a) \(x=-4\) b) \(x=-\frac{1}{2}\)

52850012

Solve each exponential equation for \(x\). Use the change-of-base formula with common logarithms, and round each answer to the nearest hundredth. a) \(2^x=50\) b) \(1.04^x=3\) c) \(0.9^x=0.5\)

Hints

- Rewrite each equation in logarithmic form. - Use the relationship between exponential and logarithmic forms. - Apply the change-of-base formula so you can evaluate the logarithm with a calculator. - Estimate first to check whether each decimal answer is reasonable.

Solution

1. For an equation of the form \(b^x=a\), use \(x=\log_b(a)=\frac{\log(a)}{\log(b)}\). 2. a) \(x=\frac{\log(50)}{\log(2)}\approx5.64\). 3. b) \(x=\frac{\log(3)}{\log(1.04)}\approx28.01\). 4. c) \(x=\frac{\log(0.5)}{\log(0.9)}\approx6.58\).

Answer

a) \(x\approx5.64\) b) \(x\approx28.01\) c) \(x\approx6.58\)

52983712

Find the base \(a\) of the exponential function \(f(x)=a^x\) if its graph passes through \((-2, 6.25)\).

Hints

- Substitute the point into the function. - Rewrite the negative exponent as a reciprocal. - Remember that an exponential base must be positive.

Solution

1. Substitute the point: \(6.25=a^{-2}\). 2. Since \(6.25=\frac{25}{4}\), \(a^2=\frac{4}{25}\). 3. An exponential base must be positive, so \(a=\frac{2}{5}=0.4\).

Answer

\(a=0.4\)

52983812

An exponential function has the form \(g(x)=b^x\). Its graph passes through \((0.75, 8)\). Find \(b\).

Hints

- Rewrite \(0.75\) as a fraction. - Raise both sides to the reciprocal of the exponent. - Use exponent rules to simplify the result.

Solution

1. Substitute the point: \(8=b^{0.75}=b^{3/4}\). 2. Raise both sides to the power \(\frac{4}{3}\): \(b=8^{4/3}\). 3. Since \(\sqrt[3]{8}=2\), \(b=2^4=16\).

Answer

\(b=16\)

52985012

The value of a new mountain bike is modeled by \(W(t)=3800(0.84)^t\), where \(W(t)\) is the value in dollars and \(t\) is the number of years since purchase. a) Find the original purchase price and the annual percent decrease in value. b) Find the value after \(4\) years. c) Find when the value decreases to one-fourth of the original purchase price.

Hints

- The coefficient is the value at \(t=0\). - Convert the remaining factor to a percent decrease. - For one-fourth of the original value, set the exponential factor equal to \(0.25\).

Solution

1. At \(t=0\), \(W(0)=3800\), so the original price was \(\$3800\). The factor \(0.84\) means the bike loses \(16\%\) of its value each year. 2. \(W(4)=3800(0.84)^4\approx 1891.91\), so the value is about \(\$1891.91\). 3. One-fourth of the initial value satisfies \((0.84)^t=0.25\). Thus, \(t=\frac{\ln(0.25)}{\ln(0.84)}\approx 7.95\) years.

Answer

a) Original price: \(\$3800\); annual decrease: \(16\%\) b) About \(\$1891.91\) c) About \(7.95\,\text{years}\)

52181312

Let \(f(x)=10(0.8)^x\). a) Describe the monotonicity and find the horizontal asymptote as \(x\to\infty\). b) A horizontal band of total width \(0.1\) is centered on the asymptote. Find the threshold \(x_1\) such that every point with \(x>x_1\) is inside the band. c) The band width is reduced to \(0.01\). Determine whether doubling the threshold from part b is sufficient to keep every point with \(x>2x_1\) inside the narrower band.

Hints

- A base between \(0\) and \(1\) produces exponential decay. - Use half the total band width. - Take logarithms and remember that dividing by a negative logarithm reverses the inequality.

Solution

1. Since \(0<0.8<1\), the function is strictly decreasing. Also, \((0.8)^x\to0\), so the horizontal asymptote is \(y=0\). 2. For a band of width \(0.1\), positive outputs must satisfy \(10(0.8)^x<0.05\). Thus, \((0.8)^x<0.005\). 3. Taking logarithms gives \(x\ln(0.8)<\ln(0.005)\). Since \(\ln(0.8)<0\), dividing reverses the inequality: \(x>\frac{\ln(0.005)}{\ln(0.8)}\approx23.74\). Hence, \(x_1\approx23.74\). 4. For width \(0.01\), require \(10(0.8)^x<0.005\), which gives \(x>\frac{\ln(0.0005)}{\ln(0.8)}\approx34.06\). 5. Since \(2x_1\approx47.48>34.06\), doubling the original threshold is sufficient.

Answer

a) Strictly decreasing; horizontal asymptote \(y=0\) b) \(x_1=\frac{\ln(0.005)}{\ln(0.8)}\approx23.74\) c) Yes. The narrower band requires only \(x>34.06\), while \(2x_1\approx47.48\).

52332412

Solve for \(x\) in each equation. a) \(\frac{1}{2}\cdot3^{x+1}=13.5\) b) \(9^{x-1}=3^{x+2}\) c) \(10\cdot2^x-15=5\)

Hints

- Rewrite a base so that both sides use the same base. - Isolate the exponential expression as you would isolate a term in a linear equation. - Use the power-of-a-power property. - Remove factors outside the exponential expression before comparing exponents.

Solution

1. a) Multiply by \(2\): \(3^{x+1}=27=3^3\). Therefore, \(x+1=3\), so \(x=2\). 2. b) Rewrite \(9\) as \(3^2\): \((3^2)^{x-1}=3^{x+2}\), so \(3^{2x-2}=3^{x+2}\). Set the exponents equal: \(2x-2=x+2\), giving \(x=4\). 3. c) Isolate the exponential term: \(10\cdot2^x=20\), so \(2^x=2=2^1\). Therefore, \(x=1\).

Answer

a) \(x=2\) b) \(x=4\) c) \(x=1\)

52379212

Solve each equation. a) \(4\cdot5^{x-1}=500\) b) \(4^{x+1}=32\) c) \(6^{2x}=1\)

Hints

- Rewrite numbers such as \(32\) as powers of a common base. - A positive base other than \(1\) produces \(1\) when the exponent is \(0\). - Isolate the exponential expression before comparing exponents. - Use the power-of-a-power property to simplify expressions.

Solution

1. a) Divide by \(4\): \(5^{x-1}=125=5^3\). Therefore, \(x-1=3\), so \(x=4\). 2. b) Rewrite both sides with base \(2\): \((2^2)^{x+1}=2^5\), so \(2^{2x+2}=2^5\). Thus, \(2x+2=5\), giving \(x=\frac{3}{2}\). 3. c) For a positive base other than \(1\), a power equals \(1\) only when its exponent is \(0\). Thus, \(2x=0\), so \(x=0\).

Answer

a) \(x=4\) b) \(x=\frac{3}{2}\) c) \(x=0\)

52456412

Solve for \(x\) in each equation. a) \(10^x=0.01\) b) \(\left(\frac{2}{5}\right)^x=6.25\) c) \(0.5^x=32\) d) \(27\cdot3^x=\frac{1}{3}\)

Hints

- Rewrite both sides as powers with the same base. - Express terminating decimals as fractions or powers of \(10\), \(2\), or \(5\). - When the bases match, set the exponents equal. - Use the product of powers property for expressions with the same base.

Solution

1. a) Since \(0.01=10^{-2}\), \(x=-2\). 2. b) Since \(6.25=\frac{25}{4}=\left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^{-2}\), \(x=-2\). 3. c) Rewrite \(0.5\) as \(2^{-1}\) and \(32\) as \(2^5\): \(2^{-x}=2^5\). Therefore, \(x=-5\). 4. d) Rewrite \(27\) as \(3^3\) and \(\frac{1}{3}\) as \(3^{-1}\): \(3^{x+3}=3^{-1}\). Thus, \(x+3=-1\), so \(x=-4\).

Answer

a) \(x=-2\) b) \(x=-2\) c) \(x=-5\) d) \(x=-4\)

52583512

Solve over the real numbers: \((x^2-4)e^x=3(x^2-4)\).

Hints

- Move all terms to one side. - Factor out the common expression. - Apply the zero-product property. - Rewrite \(e^x=3\) using a logarithm.

Solution

1. Move all terms to one side and factor: \((x^2-4)(e^x-3)=0\). 2. By the zero-product property, \(x^2-4=0\) or \(e^x-3=0\). 3. The first equation gives \(x=-2\) or \(x=2\). The second gives \(x=\ln(3)\). 4. Therefore, the solution set is \(\{-2,\ln(3),2\}\).

Answer

\(\{-2,\ln(3),2\}\)

52583612

Solve \(4^x-12\cdot2^x+32=0\).

Hints

- Express both exponential terms using base \(2\). - Substitute for \(2^x\). - Solve the resulting quadratic equation. - Back-substitute to find \(x\).

Solution

1. Rewrite \(4^x=(2^x)^2\). 2. Let \(u=2^x\). Then \(u^2-12u+32=0\). 3. Factor: \((u-4)(u-8)=0\), so \(u=4\) or \(u=8\). 4. Back-substitute: \(2^x=4\) gives \(x=2\), and \(2^x=8\) gives \(x=3\).

Answer

\(\{2,3\}\)

52610212

Find all intersection points of \(f(x)=e^{2x}-5e^x\) and \(g(x)=-6\).

Hints

- Set the function formulas equal. - Substitute for \(e^x\). - Solve the quadratic equation. - Use the constant function to determine each \(y\)-coordinate.

Solution

1. Set the functions equal: \(e^{2x}-5e^x=-6\). 2. Rearrange: \(e^{2x}-5e^x+6=0\). Let \(u=e^x\), giving \(u^2-5u+6=0\). 3. Factor: \((u-2)(u-3)=0\), so \(u=2\) or \(u=3\). 4. Thus, \(x=\ln(2)\) or \(x=\ln(3)\). Since \(g(x)=-6\), the intersection points are \((\ln(2), -6)\) and \((\ln(3), -6)\).

Answer

\((\ln(2), -6)\) and \((\ln(3), -6)\)

52620712

Solve the exponential equation. Give the solution set. \(4^{x+1}-17\cdot2^x+4=0\)

Hints

- Rewrite the power with base \(4\) using base \(2\). - Look for an expression that has quadratic form. - Substitute a new variable for \(2^x\). - After solving the quadratic, substitute back to find \(x\).

Solution

1. Rewrite \(4^{x+1}\) in terms of \(2^x\): \(4^{x+1}=4\cdot4^x=4(2^x)^2\). 2. The equation becomes \(4(2^x)^2-17\cdot2^x+4=0\). 3. Let \(u=2^x\). Then \(4u^2-17u+4=0\). 4. Factor or use the quadratic formula: \((4u-1)(u-4)=0\), so \(u=\frac{1}{4}\) or \(u=4\). 5. Substitute back. From \(2^x=\frac{1}{4}=2^{-2}\), \(x=-2\). From \(2^x=4=2^2\), \(x=2\).

Answer

\(\{-2, 2\}\)

52624512

Consider the equation \(a\cdot3^x+6=0\), where \(a\) is real. Determine the values of \(a\) for which the equation has a real solution. Justify your answer using the range of an exponential function.

Hints

- Isolate \(3^x\). - Recall the range of \(3^x\). - Determine when the right-hand side is positive. - Check the special case \(a=0\).

Solution

1. If \(a=0\), the equation becomes \(6=0\), so there is no solution. 2. For \(a\neq0\), rewrite the equation as \(3^x=-\frac{6}{a}\). 3. Since \(3^x>0\) for every real \(x\), a solution exists exactly when \(-\frac{6}{a}>0\), which occurs when \(a<0\). 4. For each \(a<0\), the exponential function is one-to-one, so there is exactly one solution: \(x=\log_3\!\left(-\frac{6}{a}\right)\).

Answer

A real solution exists exactly when \(a<0\). For \(a\geq0\), there is no real solution.

52624612

Determine whether \(4^x+4\cdot2^x+4=0\) has a real solution.

Hints

- Rewrite \(4^x\) using base \(2\). - Substitute for \(2^x\). - Remember the range of an exponential function.

Solution

1. Rewrite \(4^x=(2^x)^2\), and let \(u=2^x\). Since \(2^x>0\), \(u>0\). 2. The equation becomes \(u^2+4u+4=0\), or \((u+2)^2=0\). 3. The only quadratic solution is \(u=-2\), which contradicts \(u>0\). Therefore, the original equation has no real solution.

Answer

No real solution

52625212

Solve each equation. a) \(e^{4x}e^{2-x}=e^5\) b) \(3e^{2x-7}=\sqrt{9e}\) c) \(\frac{e^4}{e^{5x+1}}=e\sqrt{e}\)

Hints

- Combine powers with base \(e\). - Rewrite roots as rational exponents. - Compare exponents after both sides have the same base.

Solution

1. a) Combine powers: \(e^{3x+2}=e^5\). Thus, \(3x+2=5\), so \(x=1\). 2. b) Since \(\sqrt{9e}=3e^{1/2}\), divide by \(3\) and compare exponents: \(2x-7=\frac{1}{2}\). Thus, \(x=\frac{15}{4}\). 3. c) Simplify to \(e^{3-5x}=e^{3/2}\). Thus, \(3-5x=\frac{3}{2}\), so \(x=\frac{3}{10}\).

Answer

a) \(x=1\) b) \(x=\frac{15}{4}\) c) \(x=\frac{3}{10}\)

52628912

Find all zeros of \(f(x)=(x^2-2x)e^x-3e^x\).

Hints

- Factor out the common exponential term. - Recall that \(e^x\) is never \(0\). - Solve the remaining quadratic equation.

Solution

1. Factor out \(e^x\): \(f(x)=e^x(x^2-2x-3)\). 2. Since \(e^x\neq0\) for every real \(x\), the zeros come from \(x^2-2x-3=0\). 3. Factor: \((x-3)(x+1)=0\). Thus, \(x=3\) or \(x=-1\).

Answer

\(x=-1\) and \(x=3\)

52629012

Find all zeros of \(f(x)=2e^{2x}-e^{x+1}\).

Hints

- Rewrite \(e^{x+1}\) as a product. - Factor out \(e^x\). - Use the fact that an exponential expression is never zero. - Apply logarithm properties to simplify the answer.

Solution

1. Rewrite \(e^{x+1}=e^xe\) and factor: \(f(x)=e^x(2e^x-e)\). 2. Since \(e^x\neq0\), solve \(2e^x-e=0\). Then \(e^x=\frac{e}{2}\). 3. Taking natural logarithms gives \(x=\ln\!\left(\frac{e}{2}\right)=1-\ln(2)\approx0.307\).

Answer

\(x=1-\ln(2)\approx0.307\)

52630312

Solve \(3e^x-10=8e^{-x}\) exactly.

Hints

- Multiply through by \(e^x\) to remove the negative exponent. - Substitute for \(e^x\). - Check quadratic solutions against the range of \(e^x\). - Back-substitute using a logarithm.

Solution

1. Multiply by \(e^x\): \(3e^{2x}-10e^x=8\). 2. Rearrange and let \(u=e^x\): \(3u^2-10u-8=0\). 3. The quadratic formula gives \(u=4\) or \(u=-\frac{2}{3}\). 4. Since \(u=e^x>0\), discard \(-\frac{2}{3}\). From \(e^x=4\), obtain \(x=\ln(4)\).

Answer

\(x=\ln(4)\)

52630412

Find all real solutions of \(e^{2x^2}-12e^{x^2}+32=0\).

Hints

- Substitute for \(e^{x^2}\). - Solve the resulting quadratic equation. - Back-substitute and take both square roots.

Solution

1. Let \(u=e^{x^2}\). The equation becomes \(u^2-12u+32=0\). 2. Factor: \((u-4)(u-8)=0\), so \(u=4\) or \(u=8\). 3. From \(e^{x^2}=4\), obtain \(x^2=\ln(4)\), so \(x=\pm\sqrt{\ln(4)}\). 4. From \(e^{x^2}=8\), obtain \(x^2=\ln(8)\), so \(x=\pm\sqrt{\ln(8)}\).

Answer

\(\left\{-\sqrt{\ln(8)},-\sqrt{\ln(4)},\sqrt{\ln(4)},\sqrt{\ln(8)}\right\}\)

52633212

Solve \(\frac{12}{4-e^{0.5x}}=6\). Give the exact solution using the natural logarithm and an approximation rounded to three decimal places.

Hints

- Clear the fraction first. - Isolate the exponential expression. - Apply the natural logarithm. - Check that the original denominator is not zero.

Solution

1. Multiply by the denominator: \(12=6(4-e^{0.5x})\). 2. Divide by \(6\) and isolate the exponential term: \(2=4-e^{0.5x}\), so \(e^{0.5x}=2\). 3. Take natural logarithms: \(0.5x=\ln(2)\). 4. Therefore, \(x=2\ln(2)=\ln(4)\approx1.386\). The denominator is \(2\), so the solution is valid.

Answer

\(x=2\ln(2)=\ln(4)\approx1.386\)

52644712

A biology experiment tracks the area covered by a yeast culture over time. <table> <tr><td>Time \(t\) (hours)</td><td>0</td><td>4</td><td>8</td><td>12</td><td>16</td></tr> <tr><td>Area \(A(t)\) (\(\text{cm}^2\))</td><td>150</td><td>183</td><td>223</td><td>272</td><td>332</td></tr> </table> 1. Use the data to explain why an exponential growth model is appropriate. 2. Use the first and last data points to find a model of the form \(A(t)=ae^{kt}\). Round \(k\) to four decimal places.

Hints

- Compare ratios of consecutive area measurements. - The value at \(t=0\) gives the coefficient \(a\). - Substitute the last data point and use a logarithm to solve for \(k\).

Solution

1. For the equal \(4\)-hour intervals, the ratios are \(\frac{183}{150}=1.22\), \(\frac{223}{183}\approx 1.2186\), \(\frac{272}{223}\approx 1.2197\), and \(\frac{332}{272}\approx 1.2206\). Because these ratios are nearly constant, an exponential model is appropriate. 2. Since \(A(0)=150\), \(a=150\). Using \(A(16)=332\) gives \(332=150e^{16k}\). Thus, \(k=\frac{\ln(332/150)}{16}\approx 0.0497\). 3. The model is \(A(t)=150e^{0.0497t}\).

Answer

1. The ratios for consecutive equal time intervals are all approximately \(1.22\). 2. \(A(t)=150e^{0.0497t}\)

52644812

Hot tea cools in a room with constant temperature. Let \(D(t)\) be the difference, in degrees Celsius, between the tea's temperature and the room temperature after \(t\) minutes. <table> <tr><td>Time \(t\) (minutes)</td><td>0</td><td>10</td><td>20</td><td>30</td><td>40</td></tr> <tr><td>Temperature difference \(D(t)\) (\(^\circ\text{C}\))</td><td>\(72.0\)</td><td>\(54.1\)</td><td>\(40.6\)</td><td>\(30.5\)</td><td>\(22.9\)</td></tr> </table> 1. Use calculations to show that the temperature difference decreases approximately exponentially. 2. Use the first and last data points to model the cooling process with \(D(t)=D_0e^{-\lambda t}\). Give \(D_0\) and \(\lambda\), rounding \(\lambda\) to four decimal places.

Hints

- Compare ratios of consecutive temperature differences. - A nearly constant ratio over equal time intervals supports an exponential model. - Use the value at \(t=0\) for \(D_0\), then substitute the final data point and solve with logarithms.

Solution

1. For the equal \(10\)-minute intervals, the ratios are \(\frac{54.1}{72.0}\approx 0.7514\), \(\frac{40.6}{54.1}\approx 0.7505\), \(\frac{30.5}{40.6}\approx 0.7512\), and \(\frac{22.9}{30.5}\approx 0.7508\). Because these ratios are nearly constant, exponential decay is a reasonable model. 2. The initial difference is \(D_0=D(0)=72.0\). Using \(D(40)=22.9\) gives \(22.9=72.0e^{-40\lambda}\). 3. Solving, \(\lambda=-\frac{\ln(22.9/72.0)}{40}\approx 0.0286\). Therefore, \(D(t)=72.0e^{-0.0286t}\).

Answer

1. The ratios for consecutive equal time intervals are all approximately \(0.751\). 2. \(D_0=72.0\), \(\lambda\approx 0.0286\), and \(D(t)=72.0e^{-0.0286t}\)

52645812

The collectible value of a rare coin has increased by the same percent each year since 2015. In 2015 (\(t=0\)), the coin was worth \(\$120\). In 2025, it was worth \(\$195\). Its value is modeled by \(W(t)=W_0b^t\), where \(t\) is the number of years since 2015. 1. Find \(W_0\) and \(b\). Round \(b\) to four decimal places. 2. Determine when the coin's value first exceeds \(\$300\). Give the elapsed time and the first whole calendar year for which the model's yearly value exceeds \(\$300\).

Hints

- Use the value at \(t=0\) to identify \(W_0\). - Use the 2025 value to write an equation involving \(b^{10}\). - Use logarithms to solve the inequality with \(t\) in the exponent. - Distinguish the continuous threshold time from the first whole-year checkpoint.

Solution

1. The initial value is \(W_0=120\). Since 2025 is \(10\) years after 2015, \(120b^{10}=195\). Therefore, \(b^{10}=1.625\), so \(b=(1.625)^{1/10}\approx 1.0497\). 2. Solve \(120b^t>300\), or \(b^t>2.5\). Using the unrounded value of \(b\), \(t>\frac{\ln(2.5)}{\ln b}\approx 18.87\). 3. The threshold is crossed about \(18.87\) years after 2015. At whole-year checkpoints, the first year with a value above \(\$300\) is \(2015+19=2034\).

Answer

1. \(W_0=120\); \(b\approx 1.0497\) 2. About \(18.87\) years after 2015; at whole-year checkpoints, 2034

52650512

During its first years after planting, the height of a conifer is modeled by \(h(t)=\frac{3.2e^{0.5t}}{e^{0.5t}+7}\), where \(t\ge 0\) is measured in years and \(h(t)\) is measured in meters. a) According to the model, how many centimeters does the tree grow during its first \(4\) years? b) Find when the tree reaches a height of \(2.5\,\text{m}\), and verify the result by substitution.

Hints

- Find the change by subtracting the initial height from the height after \(4\) years. - Convert meters to centimeters after finding the difference. - Isolate the exponential expression before using a logarithm. - Substitute the calculated time into the model to check the height.

Solution

1. The initial height is \(h(0)=\frac{3.2}{8}=0.4\,\text{m}\). 2. After \(4\) years, \(h(4)=\frac{3.2e^2}{e^2+7}\approx 1.6433\,\text{m}\). 3. The growth is \(1.6433-0.4\approx 1.2433\,\text{m}=124.33\,\text{cm}\). 4. Set \(h(t)=2.5\): \(2.5(e^{0.5t}+7)=3.2e^{0.5t}\). This simplifies to \(0.7e^{0.5t}=17.5\), so \(e^{0.5t}=25\). 5. Therefore, \(t=2\ln 25\approx 6.44\) years. To verify, substitute \(t=2\ln 25\): then \(e^{0.5t}=25\), so \(h(t)=\frac{3.2(25)}{25+7}=2.5\,\text{m}\).

Answer

a) About \(124.33\,\text{cm}\) b) About \(6.44\,\text{years}\); substitution gives \(h(2\ln 25)=2.5\,\text{m}\)

52652912

A population is modeled by \(g(t)=120-90e^{-0.2t}\), where \(t\ge 0\) is measured in hours. a) Find all starting times \(t\) for which the population can double within some finite future interval \(T_D\). b) Find the doubling time when the starting time is \(t=0\). c) Without further calculation, explain how the required doubling time changes as the starting time \(t\) increases.

Hints

- Identify the limiting value of the function. - A doubling is possible only if twice the current value is below the limiting value. - For a start at \(t=0\), set the future value equal to twice \(g(0)\). - Consider how the growth rate changes as the function approaches its limit.

Solution

1. The population approaches but never reaches the limiting value \(120\). A future doubling is possible only when \(2g(t)<120\), or \(g(t)<60\). 2. Solve \(120-90e^{-0.2t}<60\). This gives \(e^{-0.2t}>\frac{2}{3}\), so \(t<5\ln(1.5)\approx 2.03\). Therefore, doubling is possible for \(0\le t<5\ln(1.5)\). 3. At \(t=0\), \(g(0)=30\). The doubling time satisfies \(g(T_D)=60\), so \(e^{-0.2T_D}=\frac{2}{3}\). Thus, \(T_D=5\ln(1.5)\approx 2.03\) hours. 4. As the starting time increases, the population is closer to its limiting value and its growth rate is smaller. Therefore, the time needed to double increases and approaches infinity as \(t\) approaches \(5\ln(1.5)\).

Answer

a) \(0\le t<5\ln(1.5)\approx 2.03\) b) \(T_D=5\ln(1.5)\approx 2.03\,\text{hours}\) c) The required doubling time increases as the starting time increases.

52654912

Solve each equation over the real numbers. a) \(x^2e^x=16e^x\) b) \(e^{2x}-6e^x=0\) c) \(e^x+2=15e^{-x}\)

Hints

- Factor common exponential terms. - Use that \(e^x\) is always positive. - For part c, remove the negative exponent and substitute for \(e^x\).

Solution

1. a) Factor: \(e^x(x^2-16)=0\). Since \(e^x\neq0\), \(x^2=16\), so \(x=-4\) or \(x=4\). 2. b) Factor: \(e^x(e^x-6)=0\). Thus, \(e^x=6\), so \(x=\ln(6)\). 3. c) Multiply by \(e^x\) and let \(u=e^x\): \(u^2+2u-15=0\). The solutions are \(u=3\) and \(u=-5\). Since \(u>0\), only \(u=3\) is valid, giving \(x=\ln(3)\).

Answer

a) \(\{-4,4\}\) b) \(\{\ln(6)\}\) c) \(\{\ln(3)\}\)

52655012

Solve each equation. a) \((e^{2x}-9)(x^3-8)=0\) b) \(e^{2x}-7e^x+12=0\) c) \(\frac{2x-10}{e^x}=0\)

Hints

- Use the zero-product property in part a. - Substitute for \(e^x\) in part b. - A fraction is zero when its numerator is zero and its denominator is nonzero.

Solution

1. a) By the zero-product property, \(e^{2x}=9\) or \(x^3=8\). These give \(x=\ln(3)\) or \(x=2\). 2. b) Let \(u=e^x\). Then \(u^2-7u+12=0\), so \(u=3\) or \(u=4\). Therefore, \(x=\ln(3)\) or \(x=\ln(4)\). 3. c) Since \(e^x\neq0\), the fraction equals \(0\) exactly when \(2x-10=0\). Thus, \(x=5\).

Answer

a) \(\{\ln(3),2\}\) b) \(\{\ln(3),\ln(4)\}\) c) \(\{5\}\)

52701912

Solve each exponential equation. Round when needed to three decimal places. a) \(5e^{0.5x}=25\) b) \(12(0.75)^n=6\) c) \(e^{2t+1}+10=4\)

Hints

- Isolate the exponential expression. - Take logarithms to bring down an exponent. - Check whether the isolated exponential value is positive.

Solution

1. a) Divide by \(5\): \(e^{0.5x}=5\). Thus, \(0.5x=\ln(5)\), so \(x=2\ln(5)\approx3.219\). 2. b) Divide by \(12\): \((0.75)^n=0.5\). Taking logarithms gives \(n=\frac{\ln(0.5)}{\ln(0.75)}\approx2.409\). 3. c) Isolating the exponential gives \(e^{2t+1}=-6\). Since an exponential expression is always positive, there is no real solution.

Answer

a) \(x=2\ln(5)\approx3.219\) b) \(n=\frac{\ln(0.5)}{\ln(0.75)}\approx2.409\) c) No real solution

52739712

Find the domain and all zeros of the function \(f(x)=\frac{x^2-2x-8}{e^{2x}-e^4}\).

Hints

- A quotient is defined only when its denominator is nonzero. - Solve the exponential equation that results from setting the denominator equal to \(0\). - A quotient is zero when its numerator is zero and its denominator is nonzero.

Solution

1. The denominator cannot equal \(0\). Solve \(e^{2x}-e^4=0\): \(e^{2x}=e^4\), so \(2x=4\) and \(x=2\). Therefore, the domain is \(\mathbb{R}\setminus\{2\}\). 2. A zero of the quotient occurs when the numerator is \(0\) and the denominator is not \(0\). 3. Factor the numerator: \(x^2-2x-8=(x-4)(x+2)\). The candidates are \(x=4\) and \(x=-2\). 4. Neither candidate is excluded from the domain, so both are zeros.

Answer

Domain: \(\mathbb{R}\setminus\{2\}\) Zeros: \(x=-2\) and \(x=4\)

52741712

Find the domain and all zeros of the function \(f(x)=\frac{e^{2x}-5e^x+6}{x-\ln(2)}\).

Hints

- A quotient is undefined when its denominator is \(0\) and equals \(0\) when its numerator is \(0\). - Determine whether any value makes the denominator zero. - Substitute a new variable for \(e^x\) in the numerator. - Check every candidate zero against the domain.

Solution

1. The denominator cannot equal \(0\). Since \(x-\ln(2)=0\) when \(x=\ln(2)\), the domain is \(\mathbb{R}\setminus\{\ln(2)\}\). 2. A zero of the quotient must make the numerator equal to \(0\). Solve \(e^{2x}-5e^x+6=0\). 3. Let \(u=e^x\). Then \(u^2-5u+6=0\), so \((u-2)(u-3)=0\). Thus, \(u=2\) or \(u=3\). 4. Substitute back: \(e^x=2\) gives \(x=\ln(2)\), and \(e^x=3\) gives \(x=\ln(3)\). 5. The value \(\ln(2)\) is excluded from the domain, so the only zero is \(x=\ln(3)\).

Answer

Domain: \(\mathbb{R}\setminus\{\ln(2)\}\) Zero: \(x=\ln(3)\)

52769512

A metal part at \(20\,{}^\circ\text{C}\) is placed in an oven maintained at \(220\,{}^\circ\text{C}\). After \(15\) minutes, the part's temperature is \(120\,{}^\circ\text{C}\). Assume a bounded exponential heating model. a) Find a function \(T(t)\) that models the part's temperature after \(t\) minutes. b) Find when the part reaches \(200\,{}^\circ\text{C}\).

Hints

- Identify the limiting oven temperature and the initial temperature difference. - Use the temperature after \(15\) minutes to solve for \(k\). - Isolate the exponential expression before using a logarithm.

Solution

1. A bounded heating model has the form \(T(t)=S-(S-T_0)e^{-kt}\). Here, \(S=220\) and \(T_0=20\), so \(T(t)=220-200e^{-kt}\). 2. Use \(T(15)=120\): \(120=220-200e^{-15k}\). Therefore, \(e^{-15k}=0.5\), so \(k=\frac{\ln 2}{15}\approx 0.0462\). 3. Thus, \(T(t)=220-200e^{-(\ln 2/15)t}\approx 220-200e^{-0.0462t}\). 4. Set \(T(t)=200\): \(200=220-200e^{-kt}\), so \(e^{-kt}=0.1\). Therefore, \(t=\frac{\ln 10}{k}=\frac{15\ln 10}{\ln 2}\approx 49.83\) minutes.

Answer

a) \(T(t)=220-200e^{-(\ln 2/15)t}\approx 220-200e^{-0.0462t}\) b) About \(49.83\,\text{minutes}\)

52789312

In a game, the expected winnings after \(n\) planned rounds, where \(n\) is a positive integer, are modeled by \(E(n)=20n(0.75)^n\) dollars. Determine whether two consecutive values of \(n\) produce the same expected winnings, and identify them if they exist.

Hints

- Set \(E(n)\) equal to \(E(n+1)\). - Divide out common nonzero factors. - Solve the resulting linear equation and include the next integer.

Solution

1. Set consecutive terms equal: \(E(n)=E(n+1)\). 2. Then \(20n(0.75)^n=20(n+1)(0.75)^{n+1}\). 3. Divide by \(20(0.75)^n\): \(n=0.75(n+1)\). 4. Solving gives \(0.25n=0.75\), so \(n=3\). Therefore, the consecutive values are \(3\) and \(4\). 5. Indeed, \(E(3)=E(4)=\$25.3125\).

Answer

\(n=3\) and \(n=4\)

52789512

Solve \(4^{x+2}-4^x=45\). Give the exact solution and an approximation rounded to two decimal places.

Hints

- Factor out \(4^x\). - Isolate the exponential expression. - Use the change-of-base formula.

Solution

1. Factor out \(4^x\): \(4^x(4^2-1)=45\). 2. Thus, \(15\cdot4^x=45\), so \(4^x=3\). 3. Therefore, \(x=\log_4(3)=\frac{\ln(3)}{\ln(4)}\approx0.79\).

Answer

\(x=\frac{\ln(3)}{\ln(4)}\approx0.79\)

52789612

Solve \(6^x=2^{x+3}\). Give the exact solution and an approximation rounded to two decimal places.

Hints

- Take logarithms of both sides. - Move exponent factors in front of logarithms. - Collect the terms containing \(x\).

Solution

1. Take natural logarithms: \(x\ln(6)=(x+3)\ln(2)\). 2. Collect the terms containing \(x\): \(x(\ln(6)-\ln(2))=3\ln(2)\). 3. Since \(\ln(6)-\ln(2)=\ln(3)\), \(x=\frac{3\ln(2)}{\ln(3)}=\frac{\ln(8)}{\ln(3)}\approx1.89\).

Answer

\(x=\frac{\ln(8)}{\ln(3)}\approx1.89\)

52790112

As a heated metal part cools, the difference between its temperature and the ambient temperature decreases exponentially. After \(2\) minutes, the temperature difference is \(50\,\text{K}\). After \(6\) minutes, it is \(12.5\,\text{K}\). a) Find a function of the form \(f(t)=ba^t\) that models the temperature difference after \(t\) minutes. b) Find when the temperature difference is only \(1\%\) of its value at \(t=0\).

Hints

- Write one equation for each given data point. - Divide the equations to eliminate the initial value. - Find the value at \(t=0\) before calculating \(1\%\) of it. - Use logarithms to solve for time.

Solution

1. The data give \(ba^2=50\) and \(ba^6=12.5\). Dividing gives \(a^4=\frac{12.5}{50}=0.25\), so \(a=(0.25)^{1/4}=\sqrt{0.5}\approx 0.7071\). 2. Since \(a^2=0.5\), \(b(0.5)=50\), so \(b=100\). Thus, \(f(t)=100(\sqrt{0.5})^t\). 3. One percent of \(f(0)=100\) is \(1\). Solve \(100(\sqrt{0.5})^t=1\), or \((0.5)^{t/2}=0.01\). Therefore, \(t=\frac{2\ln(0.01)}{\ln(0.5)}\approx 13.29\) minutes.

Answer

a) \(f(t)=100(\sqrt{0.5})^t\approx 100(0.7071)^t\) b) About \(13.29\,\text{minutes}\)

52790212

A bacterial culture is assumed to grow exponentially. The number of bacteria at \(t=0\) is unknown. After \(1\) hour, there are \(1200\) bacteria, and after \(4\) hours, there are \(9600\). a) Find a model of the form \(f(t)=ce^{kt}\). b) Find when the number of bacteria is \(100\) times its value at \(t=0\).

Hints

- Write an equation for each measured population. - Divide the equations to eliminate \(c\). - A hundredfold increase means \(f(t)=100f(0)\). - Use logarithms to solve for time.

Solution

1. The data give \(ce^k=1200\) and \(ce^{4k}=9600\). Dividing gives \(e^{3k}=8\), so \(k=\frac{\ln 8}{3}=\ln 2\). 2. Since \(e^k=2\), \(2c=1200\), so \(c=600\). Therefore, \(f(t)=600e^{(\ln 2)t}=600(2^t)\). 3. A hundredfold increase satisfies \(f(t)=100f(0)\). Thus, \(600(2^t)=60{,}000\), so \(2^t=100\). 4. Therefore, \(t=\frac{\ln 100}{\ln 2}\approx 6.64\) hours.

Answer

a) \(f(t)=600e^{(\ln 2)t}=600(2^t)\) b) About \(6.64\,\text{hours}\)

52817612

Consider the equations (1) \(x^4=16\) (2) \(2^x=16\) (3) \(x^{-1}=0.2\) (4) \(5^x=\sqrt[3]{25}\). a) Solve each equation over the real numbers. Include every real solution. b) Explain the main structural difference between equations (1) and (3) and equations (2) and (4). c) Explain why \(10^x=-10\) has no real solution while \(x^3=-1000\) does.

Hints

- First identify whether \(x\) is in the base or in the exponent. - For an even power, check whether both a positive and a negative value work. - Consider the range of an exponential function with a positive base.

Solution

1. a) (1) From \(x^4=16\), \(x=\pm2\). 2. (2) Since \(16=2^4\), \(x=4\). 3. (3) Since \(x^{-1}=\frac{1}{x}=\frac{1}{5}\), \(x=5\). 4. (4) Rewrite \(\sqrt[3]{25}=\sqrt[3]{5^2}=5^{2/3}\), so \(x=\frac{2}{3}\). 5. b) In (1) and (3), the variable is the base of a power. In (2) and (4), the variable is the exponent of an exponential expression. 6. c) For every real \(x\), \(10^x>0\), so it cannot equal \(-10\). An odd power can be negative; specifically, \((-10)^3=-1000\).

Answer

a) (1) \(x=-2\) or \(x=2\); (2) \(x=4\); (3) \(x=5\); (4) \(x=\frac{2}{3}\) b) In (1) and (3), \(x\) is the base. In (2) and (4), \(x\) is the exponent. c) Exponential expressions with positive bases are always positive, but odd powers can have negative outputs.

52817912

Solve each exponential equation by rewriting both sides with a common base and comparing exponents. a) \(5^{3x-2}=125\) b) \(2^{x+4}=\frac{1}{32}\) c) \(10^{x^2-5}=10{,}000\)

Hints

- Rewrite the right side as a power with the same base as the left side. - Use negative exponents to represent reciprocals. - Identify the exponent that produces \(10{,}000\) from base \(10\). - A quadratic equation in the exponent may produce two real solutions.

Solution

1. a) Rewrite \(125\) as \(5^3\). Then \(3x-2=3\), so \(3x=5\) and \(x=\frac{5}{3}\). 2. b) Rewrite \(\frac{1}{32}\) as \(2^{-5}\). Then \(x+4=-5\), so \(x=-9\). 3. c) Rewrite \(10{,}000\) as \(10^4\). Then \(x^2-5=4\), so \(x^2=9\) and \(x=-3\) or \(x=3\).

Answer

a) \(x=\frac{5}{3}\) b) \(x=-9\) c) \(x\in\{-3, 3\}\)

52818012

Solve each equation by comparing exponents. a) \(7^{4x+2}=1\) b) \(3^{2x}=\frac{1}{27}\) c) \(4^{x-3}=2^{x+1}\) d) \(a^{x^2+x}=a^2\), where \(a>0\) and \(a\ne1\)

Hints

- Identify the exponent that makes a positive base other than \(1\) equal to \(1\). - Rewrite a reciprocal using a negative exponent. - Express \(4\) as a power of \(2\). - For part d, set the exponents equal and solve the resulting quadratic equation.

Solution

1. a) Since \(1=7^0\), \(4x+2=0\), so \(x=-\frac{1}{2}\). 2. b) Since \(\frac{1}{27}=3^{-3}\), \(2x=-3\), so \(x=-\frac{3}{2}\). 3. c) Rewrite \(4\) as \(2^2\): \(2^{2x-6}=2^{x+1}\). Thus, \(2x-6=x+1\), giving \(x=7\). 4. d) Because \(a>0\) and \(a\ne1\), set the exponents equal: \(x^2+x=2\). Then \(x^2+x-2=0\), so \((x+2)(x-1)=0\). Therefore, \(x=-2\) or \(x=1\).

Answer

a) \(\left\{-\frac{1}{2}\right\}\) b) \(\left\{-\frac{3}{2}\right\}\) c) \(\{7\}\) d) \(\{-2, 1\}\)

52820812

Solve \(0.8^x=0.2\). Use the change-of-base formula and round the solution to three decimal places.

Hints

- Convert the exponential equation to logarithmic form. - Use the change-of-base formula to evaluate a logarithm with an arbitrary base. - Predict the sign of the answer by considering repeated powers of a number between \(0\) and \(1\). - Use a calculator estimate to check the size of the result.

Solution

1. Rewrite the exponential equation in logarithmic form: \(x=\log_{0.8}(0.2)\). 2. Apply the change-of-base formula: \(x=\frac{\ln(0.2)}{\ln(0.8)}\). 3. Using a calculator, \(\ln(0.2)\approx-1.609438\) and \(\ln(0.8)\approx-0.223144\). 4. Therefore, \(x\approx\frac{-1.609438}{-0.223144}\approx7.212567\). 5. Rounded to three decimal places, \(x\approx7.213\).

Answer

\(x\approx7.213\)

52821512

Approximate \(\log(5)\), where \(\log\) denotes the common logarithm, by successively narrowing an interval for the solution of \(10^x = 5\). Round to two decimal places and document each interval by checking its endpoints with powers of \(10\).

Hints

- Between which integer powers of \(10\) does \(5\) lie? - Treat the problem as solving \(10^x = 5\) by testing and narrowing values of \(x\). - The value \(10^{0.5}\) equals \(\sqrt{10}\), which provides a useful reference point. - Use a calculator to evaluate powers with decimal exponents.

Solution

1. Since \(10^0 = 1\) and \(10^1 = 10\), \(0 < \log(5) < 1\). 2. Check tenths: \(10^{0.6} \approx 3.981\) and \(10^{0.7} \approx 5.012\). Therefore, \(0.6 < \log(5) < 0.7\). 3. Check hundredths: \(10^{0.69} \approx 4.898\) and \(10^{0.70} \approx 5.012\). Therefore, \(0.69 < \log(5) < 0.70\). 4. Check thousandths to decide the rounding: \(10^{0.698} \approx 4.989\) and \(10^{0.699} \approx 5.0003\). Thus, \(0.698 < \log(5) < 0.699\). 5. Therefore, \(\log(5) \approx 0.70\) to two decimal places.

Answer

\(\log(5) \approx 0.70\)

52821612

Approximate \(\log_2(3)\) by successively narrowing an interval for the solution of \(2^x = 3\). Round to two decimal places and test decimal exponents with base \(2\).

Hints

- Rewrite \(\log_2(3) = x\) as an exponential equation. - First locate the value between two consecutive integers. - Use a calculator to test decimal exponents and compare the results with \(3\). - After finding an interval of width \(0.1\), continue narrowing it by testing values inside the interval.

Solution

1. Since \(2^1 = 2\) and \(2^2 = 4\), \(1 < \log_2(3) < 2\). 2. Check tenths: \(2^{1.5} \approx 2.828\) and \(2^{1.6} \approx 3.031\). Therefore, \(1.5 < \log_2(3) < 1.6\). 3. Check hundredths: \(2^{1.58} \approx 2.9897\) and \(2^{1.59} \approx 3.0105\). Therefore, \(1.58 < \log_2(3) < 1.59\). 4. Check thousandths: \(2^{1.584} \approx 2.9980\) and \(2^{1.585} \approx 3.0001\). Thus, \(1.584 < \log_2(3) < 1.585\). 5. The exact value is below the rounding boundary \(1.585\), so \(\log_2(3) \approx 1.58\) to two decimal places.

Answer

\(\log_2(3) \approx 1.58\)

52827112

Solve the equation and round the solution to two decimal places. \(2\cdot1.8^{x+4}=11\)

Hints

- Isolate the exponential expression first. - Apply a logarithm to bring the variable out of the exponent. - Perform the operations in order and round only at the end.

Solution

1. Divide by \(2\): \(1.8^{x+4}=5.5\). 2. Take the natural logarithm of both sides: \(\ln(1.8^{x+4})=\ln(5.5)\). 3. Apply the power property: \((x+4)\ln(1.8)=\ln(5.5)\). 4. Solve for \(x\): \(x=\frac{\ln(5.5)}{\ln(1.8)}-4\). 5. Using a calculator, \(x\approx-1.0997\), so \(x\approx-1.10\).

Answer

\(x\approx-1.10\)

52831612

Find the investment time \(n\), in years, for each case. Round to the nearest tenth. a) Initial balance: \(\$10{,}000.00\); annual rate: \(4\%\); final balance: \(\$15{,}000.00\) b) Initial balance: \(\$500.00\); annual rate: \(6\%\); final balance: \(\$1200.00\)

Hints

- Isolate the exponential expression before applying logarithms. - Use a logarithm to solve for an unknown exponent. - Apply the power property after taking logarithms.

Solution

1. a) Write \(15{,}000=10{,}000\cdot1.04^n\). Divide by \(10{,}000\): \(1.5=1.04^n\). 2. Take logarithms: \(n=\frac{\log(1.5)}{\log(1.04)}\approx10.3\) years. 3. b) Write \(1200=500\cdot1.06^n\). Divide by \(500\): \(2.4=1.06^n\). 4. Take logarithms: \(n=\frac{\log(2.4)}{\log(1.06)}\approx15.0\) years.

Answer

a) \(n\approx10.3\) years b) \(n\approx15.0\) years

52831912

A rare bird population in a wildlife preserve grows by \(6\%\) each year. At the start of the study, there are \(80\) birds. a) Find the population after \(15\) years. b) After how many years will the population have tripled?

Hints

- Convert the percent increase to a growth factor. - Write an exponential growth model using the initial population. - A tripled population is three times the initial value. - Use logarithms when the unknown appears in the exponent.

Solution

1. The growth factor is \(1.06\), so the model is \(B(t)=80\cdot1.06^t\). 2. a) \(B(15)=80\cdot1.06^{15}\approx191.72\). The model predicts about \(192\) birds. 3. b) For a tripled population, solve \(80\cdot1.06^t=3\cdot80\), so \(1.06^t=3\). 4. Therefore, \(t=\frac{\log(3)}{\log(1.06)}\approx18.85\) years, or about \(18.9\) years.

Answer

a) About \(192\) birds b) About \(18.9\) years

52832312

Two cities are growing at different rates. City A has \(25{,}000\) residents and grows by \(3.5\%\) per year. City B has \(40{,}000\) residents and grows by \(1.2\%\) per year. After how many years will the two cities have the same population, assuming the growth rates remain constant?

Hints

- Write an exponential growth model for each city. - Equal populations mean the two models have the same value. - Divide to combine the exponential expressions into one power. - Use logarithms to solve for the exponent.

Solution

1. Model the populations by \(A(t)=25{,}000\cdot1.035^t\) and \(B(t)=40{,}000\cdot1.012^t\). 2. Set the models equal: \(25{,}000\cdot1.035^t=40{,}000\cdot1.012^t\). 3. Divide to obtain \(\left(\frac{1.035}{1.012}\right)^t=1.6\). 4. Take logarithms: \(t=\frac{\ln(1.6)}{\ln(1.035)-\ln(1.012)}\approx20.91\).

Answer

After about \(20.9\) years

52833912

For exponential growth at a rate of \(p\%\) per time unit, the time \(n\) needed for a quantity to triple can be estimated by \(n\approx\frac{110}{p}\). 1. For annual rates of \(4\%\) and \(8\%\), find the exact tripling time to the nearest hundredth of a year. 2. Use the rule of thumb for each rate and find the error compared with the exact value.

Hints

- Write an exponential growth equation for a tripled quantity. - Convert each percent increase to a growth factor. - Use logarithms to solve an equation of the form \(q^n=3\). - Compare each estimate with its corresponding exact value.

Solution

1. The exact tripling time satisfies \(\left(1+\frac{p}{100}\right)^n=3\), so \(n=\frac{\ln(3)}{\ln\left(1+\frac{p}{100}\right)}\). 2. For \(p=4\), \(n\approx\frac{\ln(3)}{\ln(1.04)}\approx28.01\) years. 3. For \(p=8\), \(n\approx\frac{\ln(3)}{\ln(1.08)}\approx14.27\) years. 4. The estimates are \(\frac{110}{4}=27.50\) years and \(\frac{110}{8}=13.75\) years. 5. The errors are about \(28.01-27.50=0.51\) year and \(14.27-13.75=0.52\) year. The rule slightly underestimates both times.

Answer

1. At \(4\%\): \(28.01\) years; at \(8\%\): \(14.27\) years 2. Estimates: \(27.50\) years and \(13.75\) years; errors: about \(0.51\) year and \(0.52\) year

52835512

A window film reduces the intensity of transmitted light by \(12\%\) for each millimeter of thickness. 1. What percent of the original light intensity \(I_0\) remains after \(3\,\text{mm}\)? 2. Find the thickness, to the nearest hundredth of a millimeter, that reduces the light intensity to one half of its original value. 3. How thick must the film be so that only \(10\%\) of the original intensity is transmitted?

Hints

- Convert the percent decrease to a multiplicative factor. - Raise the factor to the number of millimeters. - Use logarithms to solve for an unknown exponent. - Write each target as an equation of the form \(q^x=y\).

Solution

1. The decay factor per millimeter is \(q=1-0.12=0.88\). 2. After \(3\,\text{mm}\), the remaining fraction is \(0.88^3=0.681472\), or about \(68.15\%\). 3. For half the original intensity, solve \(0.88^x=0.5\): \(x=\frac{\log(0.5)}{\log(0.88)}\approx5.42\,\text{mm}\). 4. For \(10\%\) transmission, solve \(0.88^x=0.1\): \(x=\frac{\log(0.1)}{\log(0.88)}\approx18.01\,\text{mm}\).

Answer

1. About \(68.15\%\) 2. About \(5.42\,\text{mm}\) 3. About \(18.01\,\text{mm}\)

52840912

Solve each equation and round each solution to two decimal places. a) \(8\cdot1.4^x=20\) b) \(15=2\cdot3^{x+1}\) c) \(0.5\cdot4^{2x}=10\)

Hints

- Isolate the exponential expression first. - Apply a logarithm to undo the exponential operation. - Use the power property to move an exponent in front of a logarithm. - After logarithming, finish isolating \(x\) when the exponent contains \(x+1\) or \(2x\).

Solution

1. a) Divide by \(8\): \(1.4^x=2.5\). Thus, \(x=\frac{\log(2.5)}{\log(1.4)}\approx2.72\). 2. b) Divide by \(2\): \(3^{x+1}=7.5\). Then \(x+1=\frac{\log(7.5)}{\log(3)}\), so \(x\approx0.83\). 3. c) Divide by \(0.5\): \(4^{2x}=20\). Then \(2x=\frac{\log(20)}{\log(4)}\), so \(x=\frac{\log(20)}{2\log(4)}\approx1.08\).

Answer

a) \(x\approx2.72\) b) \(x\approx0.83\) c) \(x\approx1.08\)

52841312

Find the coordinates of the intersection of the graphs of \(f(x)=2\cdot9^x\) and \(g(x)=54\cdot3^x\).

Hints

- At an intersection, the two function values are equal. - Rewrite both sides using the same exponential base. - Use exponent properties to combine powers. - After finding \(x\), substitute it into either function to find \(y\).

Solution

1. At an intersection, the function values are equal: \(2\cdot9^x=54\cdot3^x\). 2. Divide by \(2\): \(9^x=27\cdot3^x\). 3. Rewrite with base \(3\): \(3^{2x}=3^{x+3}\). 4. Set the exponents equal: \(2x=x+3\), so \(x=3\). 5. Substitute into either function: \(f(3)=2\cdot9^3=2\cdot729=1458\). 6. The intersection is \((3, 1458)\).

Answer

\((3, 1458)\)

52841412

The functions are \(f(x)=10\cdot1.2^x\) and \(g(x)=25\cdot0.8^x\). Find the coordinates of the intersection of their graphs. Round both coordinates to two decimal places.

Hints

- Set the two function values equal. - Combine powers with the same exponent by dividing their bases. - Use logarithms when the resulting bases cannot be matched exactly. - Use the unrounded \(x\)-value to calculate \(y\).

Solution

1. Set the function values equal: \(10\cdot1.2^x=25\cdot0.8^x\). 2. Divide to collect the exponential expressions: \(\left(\frac{1.2}{0.8}\right)^x=\frac{25}{10}\), so \(1.5^x=2.5\). 3. Solve with logarithms: \(x=\frac{\log(2.5)}{\log(1.5)}\approx2.259851\), so \(x\approx2.26\). 4. Substitute the unrounded value into either function: \(y=10\cdot1.2^{2.259851\ldots}\approx15.098640\), so \(y\approx15.10\). 5. The intersection is approximately \((2.26, 15.10)\).

Answer

\((2.26, 15.10)\)

52841512

Find the exact solution of each exponential equation. Then give a decimal approximation rounded to two decimal places. a) \(6^x=45\) b) \(2\cdot5^{x+1}=70\) c) \(4^{2x}=3\cdot8^x\)

Hints

- Isolate the exponential expression before applying logarithms. - Use logarithms to bring a variable exponent in front. - Rewrite different bases with a common base when possible. - Apply exponent properties to simplify powers with matching bases.

Solution

1. a) Apply logarithms: \(x\log(6)=\log(45)\). Thus, \(x=\frac{\log(45)}{\log(6)}\approx2.12\). 2. b) Divide by \(2\): \(5^{x+1}=35\). Therefore, \(x+1=\frac{\log(35)}{\log(5)}\), so \(x=\frac{\log(35)}{\log(5)}-1\approx1.21\). 3. c) Rewrite with base \(2\): \(2^{4x}=3\cdot2^{3x}\). Divide by \(2^{3x}\): \(2^x=3\). Hence, \(x=\frac{\log(3)}{\log(2)}\approx1.58\).

Answer

a) \(x=\frac{\log(45)}{\log(6)}\approx2.12\) b) \(x=\frac{\log(35)}{\log(5)}-1\approx1.21\) c) \(x=\frac{\log(3)}{\log(2)}\approx1.58\)

52841812

A radioactive isotope loses \(12\%\) of its amount each day. After how many days will only one fourth of the original amount remain? Round to the nearest tenth.

Hints

- Convert the percent decrease to a daily multiplier. - Express one fourth as a decimal. - The initial amount cancels when you work with the remaining fraction. - Use logarithms to isolate the exponent.

Solution

1. The daily decay factor is \(0.88\). 2. Solve \(0.88^t=0.25\). 3. Taking logarithms gives \(t=\frac{\log(0.25)}{\log(0.88)}\approx10.84\). 4. Rounded to the nearest tenth, \(t\approx10.8\) days.

Answer

About \(10.8\) days

52842012

Solve each equation for \(x\). Assume \(a>0\), \(a\ne1\), \(b>0\), and \(b\ne1\). a) \(a^{x^2-3x}=a^{x-3}\) b) \((b^{x+1})^{x-1}=b^3\)

Hints

- Which exponent property simplifies an expression of the form \((b^m)^n\)? - Look for a difference of squares in the exponent. - After setting the exponents equal, what method can you use to solve the resulting quadratic equation? - A quadratic equation can have at most how many real solutions?

Solution

1. a) Since the bases are equal, set the exponents equal: \(x^2-3x=x-3\). 2. Rewrite as \(x^2-4x+3=0\), then factor: \((x-1)(x-3)=0\). Therefore, \(x=1\) or \(x=3\). 3. b) Use the power-of-a-power property: \((b^{x+1})^{x-1}=b^{(x+1)(x-1)}=b^{x^2-1}\). 4. Set the exponents equal: \(x^2-1=3\). Then \(x^2=4\), so \(x=2\) or \(x=-2\).

Answer

a) \(x=1\) or \(x=3\) b) \(x=2\) or \(x=-2\)

52850112

Solve the exponential equation without using logarithms: \((3^x)^{x+4}=\left(\frac{1}{3}\right)^{x+6}\)

Hints

- Can you rewrite both sides as powers with the same base? - Express the reciprocal as a power with a negative exponent. - Use the power-of-a-power property on the left side. - Once the bases match, set the exponents equal.

Solution

1. Rewrite both sides with base \(3\): \((3^x)^{x+4}=3^{x(x+4)}=3^{x^2+4x}\) and \(\left(\frac{1}{3}\right)^{x+6}=3^{-x-6}\). 2. Set the exponents equal: \(x^2+4x=-x-6\). 3. Rewrite as \(x^2+5x+6=0\), then factor: \((x+2)(x+3)=0\). 4. Therefore, \(x=-2\) or \(x=-3\).

Answer

\(x=-3\) or \(x=-2\)

52850512

Solve each exponential equation. Round each answer to the nearest hundredth. a) \(3\cdot4^{x-2}=15\) b) \(5+2\cdot3^x=21\) c) \(0.5\cdot1.2^{2x}=4\)

Hints

- Isolate the exponential expression before taking a logarithm. - Taking a logarithm allows you to move an exponent in front as a factor. - Perform additions, subtractions, and divisions outside the exponential expression first. - Treat the entire exponent as one expression when applying logarithms.

Solution

1. a) Divide by \(3\): \(4^{x-2}=5\). Take logarithms: \((x-2)\ln(4)=\ln(5)\). Thus, \(x=2+\frac{\ln(5)}{\ln(4)}\approx3.16\). 2. b) Subtract \(5\) and divide by \(2\): \(3^x=8\). Therefore, \(x=\frac{\ln(8)}{\ln(3)}\approx1.89\). 3. c) Multiply by \(2\): \(1.2^{2x}=8\). Take logarithms: \(2x\ln(1.2)=\ln(8)\). Therefore, \(x=\frac{\ln(8)}{2\ln(1.2)}\approx5.70\).

Answer

a) \(x\approx3.16\) b) \(x\approx1.89\) c) \(x\approx5.70\)

52850712

Solve \(6^{x-1}=5^x\) using logarithm properties. Round the answer to the nearest hundredth.

Hints

- Take a logarithm of both sides. - Use the power property to move each exponent in front of its logarithm. - Collect all terms containing \(x\) on one side, then factor out \(x\).

Solution

1. Take common logarithms: \(\log(6^{x-1})=\log(5^x)\). 2. Apply the power property: \((x-1)\log(6)=x\log(5)\). 3. Expand and collect the terms containing \(x\): \(x(\log(6)-\log(5))=\log(6)\). 4. Therefore, \(x=\frac{\log(6)}{\log(6)-\log(5)}\approx9.83\).

Answer

\(x=\frac{\log(6)}{\log(6)-\log(5)}\approx9.83\)

52850812

Solve \(9^x=2\cdot3^{x+1}\). First use exponent properties to simplify the equation, then apply logarithms. Give the exact answer and a decimal approximation to the nearest thousandth.

Hints

- Rewrite the base \(9\) as a power of \(3\). - Express \(3^{x+1}\) as a product. - Simplify until only one exponential expression contains \(x\).

Solution

1. Rewrite \(9^x\) as \((3^2)^x=3^{2x}\) and \(3^{x+1}\) as \(3\cdot3^x\). 2. The equation becomes \(3^{2x}=6\cdot3^x\). 3. Divide by \(3^x\), which is always positive: \(3^x=6\). 4. Therefore, \(x=\log_3(6)=\frac{\log(6)}{\log(3)}\approx1.631\).

Answer

\(x=\log_3(6)\approx1.631\)

52851012

An investor compares two investments. Investment A starts with \(\$12{,}500.00\) and earns \(6.5\%\) per year. Investment B starts with \(\$20{,}000.00\) and earns \(3.8\%\) per year. After how many years will the balances be equal, and when will Investment A exceed Investment B? Round the equality time to the nearest tenth.

Hints

- Write a compound-interest model for each investment. - Set the two balances equal to find the crossover time. - Divide powers with the same exponent to combine them. - Use logarithms to isolate the exponent.

Solution

1. Model the balances by \(A(t)=12{,}500\cdot1.065^t\) and \(B(t)=20{,}000\cdot1.038^t\). 2. Set the models equal: \(12{,}500\cdot1.065^t=20{,}000\cdot1.038^t\). 3. Divide to obtain \(\left(\frac{1.065}{1.038}\right)^t=1.6\). 4. Therefore, \(t=\frac{\ln(1.6)}{\ln\left(\frac{1.065}{1.038}\right)}\approx18.30\). 5. The balances are equal after about \(18.3\) years. Since Investment A has the larger growth factor, it exceeds Investment B for \(t>18.30\).

Answer

The balances are equal after about \(18.3\) years. Investment A exceeds Investment B after that time.

52852312

An algae mat on a small lake grows by \(12\%\) each day. At the start of the observation, it covers \(4.5\,\text{m}^2\). The lake has a total area of \(1800\,\text{m}^2\). After how many whole days will the lake first be completely covered, assuming unrestricted exponential growth?

Hints

- Convert the daily percent increase to a growth factor. - Set the exponential model equal to the lake's total area. - Use logarithms to solve for the time. - Round in the direction that guarantees complete coverage.

Solution

1. The growth model is \(A(t)=4.5\cdot1.12^t\). 2. Find the boundary by solving \(4.5\cdot1.12^t=1800\). Then \(1.12^t=400\). 3. Taking logarithms gives \(t=\frac{\ln(400)}{\ln(1.12)}\approx52.87\). 4. Since the lake must be fully covered, the first whole day satisfying the condition is day \(53\).

Answer

After \(53\) days

52852412

A data center's storage use grows by \(35\%\) each year. At the beginning of \(2024\), \(450\,\text{TB}\) is in use. The total storage capacity is \(15{,}000\,\text{TB}\). In what calendar year will the data center first reach its capacity if no storage is added?

Hints

- Write an exponential growth model for storage use. - Convert the percent increase to a growth factor. - Use logarithms to solve for the number of years after \(2024\). - Convert the elapsed time to a calendar year.

Solution

1. The storage model is \(S(t)=450\cdot1.35^t\), where \(t\) is measured in years after the beginning of \(2024\). 2. Solve \(450\cdot1.35^t=15{,}000\). Then \(1.35^t=\frac{100}{3}\). 3. Taking logarithms gives \(t=\frac{\ln\left(\frac{100}{3}\right)}{\ln(1.35)}\approx11.68\). 4. This is during the calendar year \(2035\).

Answer

\(2035\)

52852512

A filtration system removes \(65\%\) of a toxin during each pass. After how many passes will the amount remaining first be less than \(0.5\%\) of the original amount?

Hints

- What fraction of the toxin remains after each pass? - Write a model for the remaining fraction after \(n\) passes. - Use logarithms to find the boundary number of passes. - Check the whole numbers immediately before and after the boundary.

Solution

1. Each pass leaves a factor of \(0.35\) of the previous amount. 2. Find the boundary by solving \(0.35^n=0.005\). 3. Taking logarithms gives \(n=\frac{\log(0.005)}{\log(0.35)}\approx5.05\). 4. After \(5\) passes, about \(0.525\%\) remains, which is still above \(0.5\%\). After \(6\) passes, about \(0.184\%\) remains. 5. Therefore, \(6\) passes are required.

Answer

\(6\) passes

52852612

A machine loses \(12\%\) of its value each year. After how many whole years will its value first be less than one third of its original purchase price?

Hints

- Convert the annual percent loss to a decay factor. - Compare the remaining value with \(\frac{1}{3}\) of the original value. - Use logarithms to find the boundary time. - Check the whole years immediately before and after the boundary.

Solution

1. The yearly decay factor is \(0.88\). 2. Find the boundary by solving \(0.88^n=\frac{1}{3}\). 3. Taking logarithms gives \(n=\frac{\log\left(\frac{1}{3}\right)}{\log(0.88)}\approx8.59\). 4. After \(8\) years, about \(35.96\%\) of the original value remains, which is still more than one third. After \(9\) years, about \(31.65\%\) remains. 5. Therefore, the value first falls below one third after \(9\) whole years.

Answer

After \(9\) years

52982512

An adult drinks a large cup of coffee containing about \(120\,\text{mg}\) of caffeine at 2:00 p.m. The body eliminates approximately \(13\%\) of the remaining caffeine each hour. a) Write a function \(K(t)\), in milligrams, for the amount of caffeine remaining \(t\) hours after the coffee is consumed. b) Find the amount of caffeine remaining at 10:00 p.m. c) For some people, more than \(15\,\text{mg}\) of caffeine may interfere with sleep. Find approximately what time the amount falls below \(15\,\text{mg}\).

Hints

- Convert the percent decrease to a remaining factor. - Count the hours from 2:00 p.m. to 10:00 p.m. - Use logarithms to solve the threshold equation. - Convert the decimal part of the elapsed time to minutes.

Solution

1. A \(13\%\) hourly decrease gives an hourly factor of \(0.87\). Thus, \(K(t)=120(0.87)^t\). 2. From 2:00 p.m. to 10:00 p.m. is \(8\) hours. Therefore, \(K(8)=120(0.87)^8\approx 39.39\,\text{mg}\). 3. Solve \(120(0.87)^t=15\). Then \((0.87)^t=0.125\), so \(t=\frac{\ln(0.125)}{\ln(0.87)}\approx 14.93\) hours. 4. About \(14\) hours \(56\) minutes after 2:00 p.m. is approximately 4:56 a.m. the next day.

Answer

a) \(K(t)=120(0.87)^t\) b) About \(39.39\,\text{mg}\) c) About 4:56 a.m. the next day

52982712

The area of a park pond covered by water lilies increases by \(40\%\) each week. At the beginning of the observation (\(t=0\)), the lilies cover \(2\,\text{m}^2\). 1. Write a function \(A(t)\) that gives the covered area after \(t\) weeks. 2. Find the covered area after \(4\) weeks and the theoretical area \(2\) weeks before the observation began. 3. Find when the area reaches \(20\,\text{m}^2\).

Hints

- Convert the percent increase to a weekly growth factor. - A negative time evaluates the same model before the chosen starting time. - Use logarithms to solve for an exponent.

Solution

1. A \(40\%\) weekly increase gives a growth factor of \(1.4\). Thus, \(A(t)=2(1.4)^t\). 2. \(A(4)=2(1.4)^4=7.6832\approx 7.68\,\text{m}^2\). Also, \(A(-2)=2(1.4)^{-2}\approx 1.02\,\text{m}^2\). 3. Solve \(2(1.4)^t=20\). Then \((1.4)^t=10\), so \(t=\frac{\ln 10}{\ln(1.4)}\approx 6.84\) weeks.

Answer

1. \(A(t)=2(1.4)^t\) 2. \(A(4)\approx 7.68\,\text{m}^2\); \(A(-2)\approx 1.02\,\text{m}^2\) 3. About \(6.84\,\text{weeks}\)

52984012

Consider an exponential function of the form \(f(x)=ca^x\). When \(x\) increases by \(3\), the function value is multiplied by \(4\). a) Find \(a\). b) Find the factor by which the function value changes when \(x\) increases by \(1.5\). c) Find the factor by which the function value changes when \(x\) decreases by \(1\). d) Show that the change factor for an increase of \(2\Delta x\) is the square of the change factor for an increase of \(\Delta x\).

Hints

- Translate the three-unit change into an equation for \(a\). - A decrease in \(x\) uses a negative exponent. - Apply the exponent rule \(a^{mn}=(a^m)^n\). - Use variables to prove the general relationship.

Solution

1. The condition gives \(a^3=4\), so \(a=\sqrt[3]{4}\approx1.5874\). 2. For an increase of \(1.5\), the factor is \(a^{1.5}=(a^3)^{1/2}=2\). 3. For a decrease of \(1\), the factor is \(a^{-1}=\frac{1}{\sqrt[3]{4}}\approx0.6300\). 4. If \(q=a^{\Delta x}\), then the factor for \(2\Delta x\) is \(a^{2\Delta x}=(a^{\Delta x})^2=q^2\).

Answer

a) \(a=\sqrt[3]{4}\approx1.5874\) b) \(2\) c) About \(0.6300\) d) \(a^{2\Delta x}=(a^{\Delta x})^2\)

52984512

The table shows the growth of a bacterial culture. The variable \(x\) is time in hours, and \(y\) is the number of bacteria in thousands. <table> <tr><td>\(x\)</td><td>0</td><td>1</td><td>2</td><td>4</td></tr> <tr><td>\(y\)</td><td>\(400\)</td><td>\(520\)</td><td>\(676\)</td><td>\(1142.44\)</td></tr> </table> a) Find a function of the form \(f(x)=ca^x\) that models the data. b) Find the number of bacteria after \(8\) hours. c) Find when the number of bacteria is five times the initial number.

Hints

- Use the value at \(x=0\) to find \(c\). - Compare the values at \(x=0\) and \(x=1\) to find \(a\). - Remember that the table values are in thousands. - Use logarithms to solve for the exponent.

Solution

1. Since \(f(0)=c=400\), \(c=400\). Using \(f(1)=520\), \(400a=520\), so \(a=1.3\). Thus, \(f(x)=400(1.3)^x\). The other table values agree with the model. 2. \(f(8)=400(1.3)^8\approx 3262.923\). Since the output is in thousands, this is about \(3{,}262{,}923\) bacteria. 3. A fivefold increase satisfies \((1.3)^x=5\). Therefore, \(x=\frac{\ln 5}{\ln(1.3)}\approx 6.13\) hours.

Answer

a) \(f(x)=400(1.3)^x\) b) About \(3262.923\) thousand, or \(3{,}262{,}923\) bacteria c) About \(6.13\,\text{hours}\)

52984712

Find the equation of an exponential function \(f(x)=ab^x\) whose graph passes through \((-1, 12)\) and \((1.5, 0.375)\).

Hints

- Substitute both points into the general model. - Solve one equation for \(a\). - Combine powers of \(b\) using exponent rules. - Raise both sides to the reciprocal of the remaining exponent.

Solution

1. Substitution gives \(12=ab^{-1}\) and \(0.375=ab^{1.5}\). 2. From the first equation, \(a=12b\). Substitute into the second: \(0.375=12b^{2.5}\). 3. Thus, \(b^{2.5}=0.03125\), so \(b=(0.03125)^{1/2.5}=0.25\). 4. Then \(a=12(0.25)=3\), so \(f(x)=3(0.25)^x\).

Answer

\(f(x)=3(0.25)^x\)

52984812

The graph of \(f(x)=ab^x\), where \(a,b>0\), passes through \((2, 5)\) and \((3, 20)\). a) Find \(a\) and \(b\). b) Another function \(g(x)=cd^x\) passes through \((2, 5)\) and \((3, 15)\). Without fully calculating \(d\), explain whether \(d\) is greater than or less than \(b\).

Hints

- Divide function values whose inputs differ by \(1\). - Use one point after finding the base. - Compare the one-step growth factors of the two functions.

Solution

1. The ratio of consecutive function values is the base: \(b=\frac{f(3)}{f(2)}=\frac{20}{5}=4\). 2. Use \(f(2)=5\): \(5=a(4)^2\), so \(a=\frac{5}{16}=0.3125\). 3. Both functions equal \(5\) at \(x=2\), but one unit later \(g(3)=15<20=f(3)\). Therefore, \(d=\frac{15}{5}<\frac{20}{5}=b\), so \(d<b\).

Answer

a) \(a=0.3125\), \(b=4\) b) \(d<b\)

52984912

A bacterial culture grows exponentially. At \(t=0\), the culture contains \(2500\) bacteria. After \(6\) hours, the population has tripled. a) Find a model of the form \(N(t)=N_0b^t\). b) Find the number of bacteria after \(15\) hours. c) Find when the population reaches \(1{,}000{,}000\).

Hints

- Translate “triples in \(6\) hours” into an equation for \(b\). - Substitute the requested time into the model. - Use logarithms to solve for a time in the exponent.

Solution

1. Since the population triples in \(6\) hours, \(b^6=3\), so \(b=3^{1/6}\approx 1.2009\). Thus, \(N(t)=2500(3^{t/6})\). 2. \(N(15)=2500(3^{15/6})\approx 38{,}971.14\), so there are about \(38{,}971\) bacteria. 3. Solve \(1{,}000{,}000=2500(3^{t/6})\). Then \(3^{t/6}=400\), so \(t=\frac{6\ln 400}{\ln 3}\approx 32.72\) hours.

Answer

a) \(N(t)=2500(3^{t/6})\), with \(b=3^{1/6}\approx 1.2009\) b) About \(38{,}971\) bacteria c) About \(32.72\,\text{hours}\)

52987912

Solve \(7\cdot3^{2x-1}=4\cdot2^{x+3}\). Give an exact answer and a decimal approximation.

Hints

- Take logarithms of both sides. - Expand logarithms of products and powers. - Collect all terms containing \(x\) on one side.

Solution

1. Take natural logarithms: \(\ln(7)+(2x-1)\ln(3)=\ln(4)+(x+3)\ln(2)\). 2. Since \(\ln(4)=2\ln(2)\), collect terms to obtain \(x(2\ln(3)-\ln(2))=5\ln(2)+\ln(3)-\ln(7)\). 3. Therefore, \(x=\frac{5\ln(2)+\ln(3)-\ln(7)}{2\ln(3)-\ln(2)}\approx1.7409\).

Answer

\(x=\frac{5\ln(2)+\ln(3)-\ln(7)}{2\ln(3)-\ln(2)}\approx1.7409\)

52988012

Solve \(2^{x+1}5^{2x}=10^{x-1}5^4\). Give an exact answer and a decimal approximation.

Hints

- Rewrite base \(10\) as \(2\cdot5\). - Combine and divide powers with matching bases. - Apply a logarithm after isolating one exponential expression.

Solution

1. Rewrite \(10^{x-1}=2^{x-1}5^{x-1}\). Then the equation becomes \(2^{x+1}5^{2x}=2^{x-1}5^{x+3}\). 2. Divide powers with the same bases: \(2^2=5^{3-x}\), so \(4=5^{3-x}\). 3. Take natural logarithms: \(\ln(4)=(3-x)\ln(5)\). 4. Therefore, \(x=3-\frac{\ln(4)}{\ln(5)}\approx2.1386\).

Answer

\(x=3-\frac{\ln(4)}{\ln(5)}\approx2.1386\)

52993512

Light intensity in a lake decreases with depth. Measurements show that the intensity decreases by \(18\%\) for every additional \(3\,\text{m}\) of depth. At the surface (\(d=0\)), the intensity is \(400\,\text{W}/\text{m}^2\). a) Find the light intensity at a depth of \(10\,\text{m}\). b) Find the depth at which the intensity is \(50\,\text{W}/\text{m}^2\). c) The model can be written as \(I(d)=I_0e^{-kd}\). Find \(k\).

Hints

- Convert the percent decrease to a remaining factor. - Express the number of \(3\)-meter intervals as \(d/3\). - Use logarithms to solve for depth. - Compare the factors over a \(3\)-meter interval in the two model forms.

Solution

1. The remaining factor for each \(3\,\text{m}\) is \(0.82\), so \(I(d)=400(0.82)^{d/3}\). 2. \(I(10)=400(0.82)^{10/3}\approx 206.43\,\text{W}/\text{m}^2\). 3. For \(I(d)=50\), solve \((0.82)^{d/3}=0.125\). Thus, \(d=\frac{3\ln(0.125)}{\ln(0.82)}\approx 31.44\,\text{m}\). 4. Since \(e^{-3k}=0.82\), \(k=-\frac{\ln(0.82)}{3}\approx 0.0662\,\text{m}^{-1}\).

Answer

a) About \(206.43\,\text{W}/\text{m}^2\) b) About \(31.44\,\text{m}\) c) \(k\approx 0.0662\,\text{m}^{-1}\)

52993612

A bacterial culture is treated with an antibiotic. The population decreases by \(15\%\) every \(20\) minutes. At \(t=0\), there are \(500{,}000\) bacteria. a) How many bacteria remain after exactly \(2\) hours? b) Find when the population has decreased to \(1\%\) of its initial value. c) Find the percent decrease per minute.

Hints

- Keep the time units consistent with the \(20\)-minute factor. - One percent remaining means the population ratio is \(0.01\). - Find the factor for one minute before converting it to a percent decrease.

Solution

1. With \(t\) measured in minutes, the model is \(N(t)=500{,}000(0.85)^{t/20}\). 2. Two hours is \(120\) minutes, so \(N(120)=500{,}000(0.85)^6\approx 188{,}575\) bacteria. 3. For \(1\%\) remaining, solve \((0.85)^{t/20}=0.01\). Thus, \(t=\frac{20\ln(0.01)}{\ln(0.85)}\approx 566.72\) minutes, or about \(9\) hours \(27\) minutes. 4. The one-minute factor is \((0.85)^{1/20}\approx 0.991907\). Therefore, the decrease is about \(0.81\%\) per minute.

Answer

a) About \(188{,}575\) bacteria b) About \(566.72\,\text{minutes}\), or about \(9\) hours \(27\) minutes c) About \(0.81\%\) per minute

53002612

Two bacterial cultures are being studied in a laboratory. Culture A doubles every \(5\) hours. Culture B doubles every \(8\) hours but begins with four times as many bacteria as Culture A. Find when the two cultures will have the same number of bacteria.

Hints

- Write an exponential model for each culture. - Use a variable for Culture A's unknown initial population. - Rewrite \(4\) as a power of \(2\). - Once both sides have the same base, compare the exponents.

Solution

1. Let Culture A begin with \(N_0\) bacteria. Then \(N_A(t)=N_0 2^{t/5}\) and \(N_B(t)=4N_0 2^{t/8}\). 2. Set the populations equal: \(N_0 2^{t/5}=4N_0 2^{t/8}\). Dividing by \(N_0\) gives \(2^{t/5}=4\cdot 2^{t/8}\). 3. Since \(4=2^2\), \(2^{t/5}=2^{2+t/8}\), so \(\frac{t}{5}=2+\frac{t}{8}\). 4. Solving gives \(\frac{3t}{40}=2\), so \(t=\frac{80}{3}=26\frac{2}{3}\) hours.

Answer

The cultures have the same population after \(26\frac{2}{3}\) hours, or about \(26.67\) hours.

53003112

The population of a rapidly growing metropolitan area was \(450{,}000\) in 2005 and \(585{,}000\) in 2020. Assume the population follows an exponential growth model. a) Find the average annual growth rate from 2005 to 2020. b) Predict the population in 2045 if the same growth rate continues. c) Determine during which year the population would first exceed one million.

Hints

- Find the number of years between the two known populations. - Use \(P(t)=P_0q^t\). - The annual percent growth rate is \((q-1)\cdot 100\%\). - Use logarithms to solve for the time when the model reaches a target population.

Solution

1. The time from 2005 to 2020 is \(15\) years. If \(P(t)=450{,}000q^t\), then \(585{,}000=450{,}000q^{15}\). Therefore, \(q=(1.3)^{1/15}\approx 1.01764\), so the average annual growth rate is about \(1.76\%\). 2. The year 2045 is \(40\) years after 2005. Thus, \(P(40)=450{,}000q^{40}\approx 905{,}861\), or about \(905{,}900\) people. 3. Solve \(1{,}000{,}000=450{,}000q^t\). Then \(t=\frac{\ln(1{,}000{,}000/450{,}000)}{\ln q}\approx 45.65\). This is about \(45.65\) years after 2005, so the population first exceeds one million during 2050.

Answer

a) About \(1.76\%\) per year, with \(q\approx 1.01764\) b) About \(905{,}900\) people c) During 2050

53230012

A cup of hot cocoa starts at \(80\,{}^\circ\text{C}\) in a room maintained at \(20\,{}^\circ\text{C}\). Its temperature is modeled by \(T(t)=20+60e^{-kt}\), where \(t\) is the number of minutes since the measurement began. After \(10\) minutes, the cocoa is \(44\,{}^\circ\text{C}\). a) Find the cooling constant \(k\), rounded to three decimal places. b) Find when the cocoa reaches \(25\,{}^\circ\text{C}\). Round to the nearest tenth of a minute. c) Use the graph to estimate the temperature after \(15\) minutes. Then check the estimate using the rounded value of \(k\) from part a).

Figure for problem 532300

Hints

- Substitute the known time and temperature into the model. - Isolate the exponential expression before taking a natural logarithm. - For part b, replace \(T(t)\) with the target temperature. - Pay attention to the scales on both graph axes.

Solution

1. Use \(T(10)=44\): \(20+60e^{-10k}=44\). Then \(e^{-10k}=0.4\), so \(k=-\frac{\ln(0.4)}{10}\approx 0.092\). 2. For \(T(t)=25\), \(20+60e^{-kt}=25\), so \(e^{-kt}=\frac{1}{12}\). Using \(k=-\frac{\ln(0.4)}{10}\), \(t=\frac{10\ln(12)}{-\ln(0.4)}\approx 27.1\) minutes. 3. The graph gives an estimate of about \(35\,{}^\circ\text{C}\). Using the rounded value of \(k\) from part a), \(T(15)=20+60e^{-0.092(15)}\approx 35.1\,{}^\circ\text{C}\).

Answer

a) \(k\approx 0.092\) b) About \(27.1\) minutes c) Graph estimate: about \(35\,{}^\circ\text{C}\); calculated value: about \(35.1\,{}^\circ\text{C}\)

53267712

A metal sphere is moved from a room into an oven maintained at \(200\,{}^\circ\text{C}\). Its temperature is modeled by \(T(t)=A-Be^{-kt}\), where \(T(t)\) is measured in degrees Celsius and \(t\) is measured in minutes. The graph shows the temperature curve. a) Use the context and the graph to determine \(A\) and \(B\). b) The point \(P(10, 110)\) lies on the graph. Use it to determine \(k\), rounded to four decimal places. c) Find when the sphere reaches \(185\,{}^\circ\text{C}\). Round to the nearest tenth of a minute.

Figure for problem 532677

Hints

- The limiting temperature determines the horizontal asymptote. - Use the vertical intercept to find the initial temperature. - Substitute a known point into the model to solve for the exponent parameter. - Use a natural logarithm after isolating the exponential expression.

Solution

1. The sphere's temperature approaches the oven temperature, so \(A=200\). The graph shows \(T(0)=20\). Therefore, \(200-B=20\), so \(B=180\). 2. Use \(P(10, 110)\): \(110=200-180e^{-10k}\). Then \(e^{-10k}=0.5\), so \(k=\frac{\ln 2}{10}\approx 0.0693\). 3. Using \(k\approx 0.0693\), solve \(185=200-180e^{-0.0693t}\). This gives \(e^{-0.0693t}=\frac{1}{12}\), so \(t=\frac{\ln 12}{0.0693}\approx 35.9\) minutes.

Answer

a) \(A=200\), \(B=180\) b) \(k\approx 0.0693\) c) About \(35.9\) minutes

53402712

A new smartphone loses value quickly. The graph shows its estimated resale value \(W(t)\), in dollars, after \(t\) years. a) Use the graph to find the initial value and annual factor \(b\). Write the model in the form \(W(t)=ab^t\). b) By what percent does the value decrease each year? c) When is the phone worth one-fourth of its original price? Calculate the time or estimate it from the graph.

Figure for problem 534027

Hints

- Read the y-intercept to find the initial value. - Compare the values at \(t=0\) and \(t=1\). - Relate the decay factor to the percent loss. - One-fourth of the initial value is \(\$200\).

Solution

1. The graph shows \(W(0)=800\), so the initial value is \(a=800\). 2. The graph shows \(W(1)=600\). Therefore, \(b=\frac{600}{800}=0.75\), and the model is \(W(t)=800(0.75)^t\). 3. The yearly percent decrease is \((1-0.75)\cdot100\%=25\%\). 4. One-fourth of \(800\) is \(200\). Solve \(800(0.75)^t=200\): \(t=\frac{\ln(0.25)}{\ln(0.75)}\approx4.82\).

Answer

a) Initial value: \(\$800\); factor: \(0.75\); \(W(t)=800(0.75)^t\) b) \(25\%\) decrease per year c) About \(4.82\) years

53448212

The height of a certain pine tree is modeled by \(h(t)=12\left(1-0.75e^{-0.15t}\right)\), where \(t\) is the number of years since planting and \(h(t)\) is the height in meters. a) Find the tree's height when it is planted. b) Find when the tree reaches a height of \(9\,\text{m}\). c) Use the graph to identify the theoretical height the tree approaches but does not exceed. Briefly justify this value from the function.

Figure for problem 534482

Hints

- Planting corresponds to \(t=0\). - Isolate the exponential expression before using a logarithm. - Consider what happens to \(e^{-0.15t}\) as \(t\) becomes very large. - Look for the horizontal value approached by the graph.

Solution

1. At planting, \(t=0\). Thus, \(h(0)=12(1-0.75)=3\), so the initial height is \(3\,\text{m}\). 2. Solve \(12\left(1-0.75e^{-0.15t}\right)=9\). Then \(e^{-0.15t}=\frac{1}{3}\), so \(t=\frac{\ln 3}{0.15}\approx 7.32\) years. 3. As \(t\to\infty\), \(e^{-0.15t}\to0\), so \(h(t)\to12\). The graph approaches the horizontal line \(h=12\).

Answer

a) \(3\,\text{m}\) b) About \(7.32\) years c) \(12\,\text{m}\)

53449012

The graph shows two quantities, \(A\) and \(B\), changing over time in a biological experiment. a) Find an equation of the form \(f(t)=f(0)e^{kt}\) for each graph. b) Find the exact time when the two quantities are equal.

Figure for problem 534490

Hints

- The vertical intercept gives each initial value. - Use a clearly readable point on each graph to find \(k\). - Set the two models equal and use exponent rules before taking logarithms.

Solution

1. For quantity \(A\), the graph gives \(A(0)=20\) and \(A(2)=40\). Thus, \(40=20e^{2k_A}\), so \(k_A=\frac{\ln 2}{2}\). Therefore, \(A(t)=20e^{(\ln 2/2)t}\). 2. For quantity \(B\), the graph gives \(B(0)=80\) and \(B(2)=40\). Thus, \(40=80e^{2k_B}\), so \(k_B=-\frac{\ln 2}{2}\). Therefore, \(B(t)=80e^{-(\ln 2/2)t}\). 3. Set the models equal: \(20e^{(\ln 2/2)t}=80e^{-(\ln 2/2)t}\). This simplifies to \(e^{(\ln 2)t}=4\). Taking natural logarithms gives \((\ln 2)t=\ln 4=2\ln 2\), so \(t=2\).

Answer

a) \(A(t)=20e^{(\ln 2/2)t}\) and \(B(t)=80e^{-(\ln 2/2)t}\) b) \(t=2\)

53449112

The graph models the growth of a bacterial culture. a) Use the marked points to find a model of the form \(f(t)=ce^{kt}\). b) Find when the culture reaches \(1000\) units.

Figure for problem 534491

Hints

- Use the value at \(t=0\) to find \(c\). - Substitute another marked point to find \(k\). - Use a natural logarithm to solve for time.

Solution

1. The y-intercept gives \(c=100\). 2. Using the point \((10, 200)\), \(200=100e^{10k}\). Thus, \(e^{10k}=2\), so \(k=\frac{\ln2}{10}\approx0.0693\). Therefore, \(f(t)=100e^{(\ln2/10)t}\). 3. Set \(f(t)=1000\): \(1000=100e^{(\ln2/10)t}\). Then \(t=\frac{10\ln10}{\ln2}\approx33.22\).

Answer

a) \(f(t)=100e^{(\ln2/10)t}\approx100e^{0.0693t}\) b) \(t=\frac{10\ln10}{\ln2}\approx33.22\) time units

53460512

An empty storage reservoir is filled by a constant inflow. At the same time, water is lost through evaporation and seepage at a rate proportional to the amount of water present. The water volume \(W\), in cubic meters, approaches a limiting value. Use the graph to find a suitable model of the form \(W(t)=S\left(1-e^{-kt}\right)\), where \(t\) is measured in hours.

Figure for problem 534605

Hints

- The horizontal asymptote gives the limiting value \(S\). - Use a point that can be read accurately from the graph. - Isolate the exponential expression and take a natural logarithm.

Solution

1. The graph approaches \(500\,\text{m}^3\), so \(S=500\). 2. The graph passes through \((2,250)\). Substitute this point into \(W(t)=500\left(1-e^{-kt}\right)\): \(250=500\left(1-e^{-2k}\right)\). 3. Then \(e^{-2k}=0.5\), so \(k=\frac{\ln 2}{2}\approx 0.3466\). 4. A suitable model is \(W(t)=500\left(1-e^{-(\ln 2/2)t}\right)\), or approximately \(W(t)=500\left(1-e^{-0.3466t}\right)\).

Answer

\(W(t)=500\left(1-e^{-(\ln 2/2)t}\right)\), or approximately \(W(t)=500\left(1-e^{-0.3466t}\right)\)

52181412

Let \(g(x)=4-3(2^x)\). a) Find \(\lim_{x\to-\infty}g(x)\) and the horizontal asymptote. b) A horizontal band of total width \(0.02\) is centered on the asymptote. Find the greatest threshold \(x_0\) such that every graph point with \(x<x_0\) is inside the band. c) The band width is reduced to one fourth of its original value, \(0.005\). By how much does the threshold change?

Hints

- Determine what happens to \(2^x\) as \(x\to-\infty\). - Use half the total band width as the allowed distance. - Compare the two logarithmic thresholds using the factor of \(\frac{1}{4}\).

Solution

1. As \(x\to-\infty\), \(2^x\to0\), so \(g(x)\to4\). The horizontal asymptote is \(y=4\). 2. For width \(0.02\), the condition is \(|g(x)-4|<0.01\). Thus, \(3(2^x)<0.01\), so \(2^x<\frac{0.01}{3}\). 3. Taking logarithms gives \(x<\frac{\ln(0.01/3)}{\ln(2)}\approx-8.23\). Therefore, \(x_0\approx-8.23\). 4. For width \(0.005\), require \(3(2^x)<0.0025\), giving \(x<\frac{\ln(0.0025/3)}{\ln(2)}\approx-10.23\). 5. The new threshold is exactly \(2\) units smaller because the allowed distance was divided by \(4\), and \(\log_2\left(\frac{1}{4}\right)=-2\).

Answer

a) The limit is \(4\); horizontal asymptote \(y=4\). b) \(x_0=\frac{\ln(0.01/3)}{\ln(2)}\approx-8.23\) c) The threshold decreases by \(2\).

52552512

Consider the family of functions \(f_t(x)=e^{2x}-te^x+4\), where \(t\in\mathbb{R}\). a) Find the value of \(t\) for which the graph passes through \((0, 1)\). b) Find all values of \(t\) for which \(f_t\) has no real zeros.

Hints

- Substitute the coordinates of the given point into the function. - Use the substitution \(u=e^x\). - Remember that \(e^x\) is always positive. - Determine when the resulting quadratic has no positive roots.

Solution

1. Substitute \((0, 1)\): \(1-t+4=1\), so \(t=4\). 2. To analyze zeros, let \(u=e^x\). Since \(u>0\), the equation becomes \(u^2-tu+4=0\). 3. If \(-4<t<4\), the discriminant \(t^2-16\) is negative, so there are no real values of \(u\). 4. If \(t\leq-4\), the quadratic has real root(s) with product \(4>0\) and sum \(t<0\). Therefore, every real root is negative and cannot equal \(e^x\). 5. At \(t=4\), the equation has the positive root \(u=2\), and for \(t>4\), it has positive roots. Therefore, \(f_t\) has no real zeros exactly when \(t<4\).

Answer

a) \(t=4\) b) \(t<4\)

52620812

Solve the equation for \(x\). \(2\cdot4^x+4^{x-1}=3^{x+1}+3^x\)

Hints

- Factor a common exponential expression from each side. - Collect the terms with variable exponents into a single ratio. - Use exponent properties to combine quotients of powers. - Rewrite the constant ratio as a power of \(\frac{4}{3}\).

Solution

1. Factor the smallest power on each side: \(4^{x-1}(2\cdot4+1)=3^x(3+1)\). 2. Simplify: \(9\cdot4^{x-1}=4\cdot3^x\). 3. Since \(4^{x-1}=\frac{4^x}{4}\), multiply by \(4\) to obtain \(9\cdot4^x=16\cdot3^x\). 4. Divide by \(9\cdot3^x\): \(\left(\frac{4}{3}\right)^x=\frac{16}{9}=\left(\frac{4}{3}\right)^2\). 5. Equal powers with the same valid base have equal exponents, so \(x=2\).

Answer

\(x=2\)

52649412

A new social media channel follows the bounded-growth model \(F(t) = S - ce^{-kt}\), where \(t\) is the number of months since the channel was created. The channel has \(400\) followers at \(t=0\), \(760\) followers after \(5\) months, and \(976\) followers after \(10\) months. Find the saturation level \(S\), which represents the maximum expected number of followers.

Hints

- What value does \(F(t)\) approach as \(t\) increases? - Use each of the three data points to form an equation involving the unknown parameters. - Try substituting a variable for \(e^{-5k}\). - How is \(e^{-10k}\) related to \(e^{-5k}\)?

Solution

1. Use the initial value: \(F(0)=S-c=400\), so \(c=S-400\). 2. Let \(u=e^{-5k}\). Then \(e^{-10k}=u^2\). The data at \(t=5\) give \(S-(S-400)u=760\), so \(u=\frac{S-760}{S-400}\). 3. The data at \(t=10\) give \(S-(S-400)u^2=976\). Substitute the expression for \(u\): \(S-(S-400)\left(\frac{S-760}{S-400}\right)^2=976\). 4. Simplify: \((S-760)^2=(S-976)(S-400)\). 5. Expanding gives \(S^2-1520S+577{,}600=S^2-1376S+390{,}400\). Thus \(144S=187{,}200\), so \(S=1300\).

Answer

The saturation level is \(1300\) followers.

52653012

A population is modeled by \(f(t)=40-30e^{-0.1t}\), where \(t\ge 0\) is measured in weeks. a) Find all starting times \(t\) for which the population can increase by \(50\%\) within a finite future interval \(T\). b) Write a function for the required duration \(T\) in terms of the starting time \(t\). c) Find the minimum duration required for a \(50\%\) increase.

Hints

- Compare \(1.5f(t)\) with the limiting value of the function. - Use \(f(t+T)=1.5f(t)\). - Split \(e^{-0.1(t+T)}\) using exponent rules. - Determine where the duration function has its smallest value.

Solution

1. The limiting value is \(40\). A \(50\%\) increase is possible only if \(1.5f(t)<40\), or \(f(t)<\frac{80}{3}\). 2. Solving \(40-30e^{-0.1t}<\frac{80}{3}\) gives \(e^{-0.1t}>\frac{4}{9}\). Therefore, \(0\le t<10\ln(2.25)\approx 8.11\). 3. The required duration satisfies \(f(t+T)=1.5f(t)\). Substitution gives \(40-30e^{-0.1t}e^{-0.1T}=60-45e^{-0.1t}\). 4. Solving for the exponential term gives \(e^{-0.1T}=1.5-\frac{2}{3}e^{0.1t}\). Thus, \(T(t)=-10\ln\left(1.5-\frac{2}{3}e^{0.1t}\right)\). 5. The duration increases with \(t\), so its minimum occurs at \(t=0\). Therefore, \(T_{\min}=-10\ln\left(\frac{5}{6}\right)=10\ln(1.2)\approx 1.82\) weeks.

Answer

a) \(0\le t<10\ln(2.25)\approx 8.11\) b) \(T(t)=-10\ln\left(1.5-\frac{2}{3}e^{0.1t}\right)\) c) \(T_{\min}=10\ln(1.2)\approx 1.82\,\text{weeks}\)

52789412

Let \(E(n)=np^n\), where \(n\) is a positive integer and \(0<p<1\). Prove the claim: “There is no value of \(p\) for which three consecutive values of \(n\) have the same expected value.”

Hints

- Write the condition for two consecutive terms to be equal. - Solve each consecutive-pair condition for \(p\). - Set the two expressions for \(p\) equal and simplify.

Solution

1. The condition \(E(n)=E(n+1)\) gives \(np^n=(n+1)p^{n+1}\), so \(p=\frac{n}{n+1}\). 2. The condition \(E(n+1)=E(n+2)\) gives \((n+1)p^{n+1}=(n+2)p^{n+2}\), so \(p=\frac{n+1}{n+2}\). 3. If three consecutive values were equal, then \(\frac{n}{n+1}=\frac{n+1}{n+2}\). 4. Cross-multiplying gives \(n(n+2)=(n+1)^2\), or \(n^2+2n=n^2+2n+1\), which is impossible. 5. Therefore, no such value of \(p\) exists.

Answer

The claim is true; the required conditions lead to the contradiction \(0=1\).

52827212

Solve the equation and round the solution to two decimal places. \(5^{x-1}=2\cdot3^{x+1}\)

Hints

- Apply a logarithm to both sides. - Use the product and power properties of logarithms. - Move all terms containing \(x\) to one side. - Factor out \(x\) before dividing.

Solution

1. Take the natural logarithm of both sides: \(\ln(5^{x-1})=\ln(2\cdot3^{x+1})\). 2. Apply the product and power properties: \((x-1)\ln(5)=\ln(2)+(x+1)\ln(3)\). 3. Collect the terms containing \(x\): \(x\ln(5)-x\ln(3)=\ln(2)+\ln(3)+\ln(5)\). 4. Factor and combine logarithms: \(x(\ln(5)-\ln(3))=\ln(30)\). 5. Therefore, \(x=\frac{\ln(30)}{\ln(5/3)}\approx6.6582\), so \(x\approx6.66\).

Answer

\(x\approx6.66\)

52832012

A laptop costs \(\$2400.00\) when new and loses \(25\%\) of its value each year. a) Find its value after \(3\) years. b) At the end of which whole year will its value first be less than \(\$200.00\)?

Hints

- A \(25\%\) loss means multiplying by what factor each year? - Identify the initial value and the threshold value. - Solve the corresponding exponential equation for the boundary time. - Check the whole years immediately before and after that time.

Solution

1. The yearly decay factor is \(0.75\), so \(W(t)=2400\cdot0.75^t\). 2. a) \(W(3)=2400\cdot0.75^3=1012.50\). The value is \(\$1012.50\). 3. b) Find the boundary time by solving \(2400\cdot0.75^t=200\). Then \(0.75^t=\frac{1}{12}\). 4. Taking logarithms gives \(t=\frac{\log\left(\frac{1}{12}\right)}{\log(0.75)}\approx8.64\). 5. At the end of year \(8\), the value is about \(\$240.27\); at the end of year \(9\), it is about \(\$180.20\). Therefore, the value first falls below \(\$200.00\) at the end of year \(9\).

Answer

a) \(\$1012.50\) b) At the end of year \(9\)

52832412

Two forestry companies compare their timber inventories. Company A has \(800\,\text{m}^3\) of timber and grows by \(4\%\) per year. Company B has \(2000\,\text{m}^3\) and grows by \(1.5\%\) per year. After how many years will Company A's inventory be exactly \(75\%\) of Company B's inventory?

Hints

- Write an exponential growth model for each inventory. - Express “\(75\%\) of Company B” as a multiplication by \(0.75\). - Set Company A's inventory equal to that amount. - Use logarithms to solve for the exponent.

Solution

1. Model the inventories by \(A(t)=800\cdot1.04^t\) and \(B(t)=2000\cdot1.015^t\). 2. Set \(A(t)\) equal to \(75\%\) of \(B(t)\): \(800\cdot1.04^t=0.75(2000\cdot1.015^t)\). 3. Simplify and divide: \(\left(\frac{1.04}{1.015}\right)^t=\frac{1500}{800}=1.875\). 4. Take logarithms: \(t=\frac{\ln(1.875)}{\ln(1.04)-\ln(1.015)}\approx25.83\).

Answer

After about \(25.8\) years

52834012

Investigate the time \(n\) needed for a quantity growing at an annual rate of \(p\%\) to quadruple. 1. Find the quadrupling time for \(p=2\), \(p=5\), and \(p=10\). Round to the nearest hundredth. 2. For each case, multiply the time \(n\) by the percent rate \(p\). What approximate constant \(c\) do you obtain? Use it to write a rule of thumb in the form \(n\approx\frac{c}{p}\).

Hints

- Write an equation for a quantity that has become four times its initial value. - Use logarithms to find each time. - Examine whether the products \(np\) are approximately constant. - Choose a simple, memorable value for the constant in the rule of thumb.

Solution

1. The quadrupling time is \(n=\frac{\ln(4)}{\ln\left(1+\frac{p}{100}\right)}\). 2. For \(p=2\), \(n\approx70.01\) years and, using the unrounded time, \(np\approx140.01\). 3. For \(p=5\), \(n\approx28.41\) years and, using the unrounded time, \(np\approx142.07\). 4. For \(p=10\), \(n\approx14.55\) years and, using the unrounded time, \(np\approx145.45\). 5. The products are all near \(140\) to \(145\). One simple rule of thumb is \(n\approx\frac{140}{p}\); a constant near \(142\) would also fit these three values.

Answer

1. \(70.01\) years, \(28.41\) years, and \(14.55\) years 2. Using unrounded times, the products \(np\) are about \(140.01\), \(142.07\), and \(145.45\). A reasonable rule is \(n\approx\frac{140}{p}\).

52835012

Air pressure decreases approximately exponentially as altitude increases. For each \(1000\,\text{m}\) increase in altitude, air pressure decreases by about \(12\%\). a) What percent of sea-level air pressure remains at an altitude of \(3000\,\text{m}\)? b) Find the pressure factor for altitude increases of \(1000\,\text{m}\), \(500\,\text{m}\), and \(5000\,\text{m}\). c) At approximately what altitude is the air pressure one-half of its sea-level value?

Hints

- Write a pressure model using altitude measured in thousands of meters. - A \(500\,\text{m}\) increase is one-half of the standard \(1000\,\text{m}\) step. - For the half-pressure altitude, set the remaining fraction equal to \(0.5\).

Solution

1. The factor per \(1000\,\text{m}\) is \(b=1-0.12=0.88\). 2. At \(3000\,\text{m}\), the remaining fraction is \((0.88)^3=0.681472\), or about \(68.1\%\). 3. The factors are \(0.88\) for \(1000\,\text{m}\), \((0.88)^{0.5}=\sqrt{0.88}\approx0.9381\) for \(500\,\text{m}\), and \((0.88)^5\approx0.5277\) for \(5000\,\text{m}\). 4. Let \(x\) be the altitude in thousands of meters. Solve \((0.88)^x=0.5\): \(x=\frac{\ln(0.5)}{\ln(0.88)}\approx5.4223\). Therefore, the altitude is about \(5422\,\text{m}\).

Answer

a) About \(68.1\%\) b) \(1000\,\text{m}\): \(0.88\); \(500\,\text{m}\): about \(0.9381\); \(5000\,\text{m}\): about \(0.5277\) c) About \(5422\,\text{m}\)

52835312

Hot cocoa is placed in a room at a constant temperature of \(20\,\text{°C}\). The cocoa begins at \(80\,\text{°C}\). Each minute, the temperature difference between the cocoa and the room decreases by \(12\%\). a) Write a function \(f(x)\) for the temperature difference after \(x\) minutes. b) Find the temperature difference after \(5\) minutes and after \(15\) minutes. c) Find the cocoa's actual temperature after \(10\) minutes. d) When does the temperature difference first become less than \(5\,\text{°C}\)? Give the continuous-time result and the first whole-number minute.

Hints

- First find the initial difference between the cocoa and room temperatures. - Convert the percent decrease to a decay factor. - Distinguish the temperature difference from the cocoa's actual temperature. - Set the difference equal to \(5\) to locate the threshold time.

Solution

1. The initial difference is \(80-20=60\,\text{°C}\), and the decay factor is \(0.88\). Thus \(f(x)=60(0.88)^x\). 2. \(f(5)\approx31.66\,\text{°C}\), and \(f(15)\approx8.82\,\text{°C}\). 3. After \(10\) minutes, the difference is \(f(10)\approx16.71\,\text{°C}\), so the cocoa's temperature is \(20+16.71=36.71\,\text{°C}\). 4. Solve \(60(0.88)^x<5\). The boundary value is \(x=\frac{\ln(5/60)}{\ln(0.88)}\approx19.44\). Therefore, the difference is below \(5\,\text{°C}\) after about \(19.44\) minutes, and first at a whole-number measurement after \(20\) minutes.

Answer

a) \(f(x)=60(0.88)^x\) b) After \(5\) minutes: about \(31.66\,\text{°C}\); after \(15\) minutes: about \(8.82\,\text{°C}\) c) About \(36.71\,\text{°C}\) d) After about \(19.44\) minutes; first whole-number minute: \(20\)

52835712

A medication is modeled as decreasing in the bloodstream by \(15\%\) each hour. Immediately after a dose, \(400\,\text{mg}\) is present. a) Find the modeled amount after \(2\), \(5\), and \(10\) hours. b) When does the modeled amount reach \(100\,\text{mg}\)? c) When does the modeled amount reach one-tenth of the initial amount?

Hints

- What decay factor represents a \(15\%\) hourly decrease? - Write an exponential-decay function. - Decide which parts require evaluating the function and which require solving for the exponent. - Use logarithms when the unknown is in the exponent.

Solution

1. The decay factor is \(0.85\), so \(A(t)=400(0.85)^t\). 2. \(A(2)=289\,\text{mg}\), \(A(5)\approx177.48\,\text{mg}\), and \(A(10)\approx78.75\,\text{mg}\). 3. For \(100\,\text{mg}\), solve \(400(0.85)^t=100\). Then \(t=\frac{\ln(0.25)}{\ln(0.85)}\approx8.53\) hours. 4. One-tenth of the initial amount is \(40\,\text{mg}\). Solving \(400(0.85)^t=40\) gives \(t=\frac{\ln(0.1)}{\ln(0.85)}\approx14.17\) hours.

Answer

a) \(2\) hours: \(289\,\text{mg}\); \(5\) hours: about \(177.48\,\text{mg}\); \(10\) hours: about \(78.75\,\text{mg}\) b) About \(8.53\) hours c) About \(14.17\) hours

52841012

Solve each exponential equation. Round each solution to two decimal places. a) \(5\cdot2^x=3^{x+2}\) b) \(6^{2x-1}=12\cdot4^x\)

Hints

- Use exponent properties to separate constant powers from powers containing \(x\). - Move all exponential expressions with \(x\) into one ratio. - Combine powers with the same exponent as \(\frac{a^x}{b^x}=\left(\frac{a}{b}\right)^x\). - Rewrite \(6^{2x}\) as \(36^x\) to align the exponents.

Solution

1. a) Rewrite \(3^{x+2}\) as \(9\cdot3^x\). Then \(5\cdot2^x=9\cdot3^x\), so \(\left(\frac{2}{3}\right)^x=\frac{9}{5}=1.8\). 2. Therefore, \(x=\frac{\log(1.8)}{\log(2/3)}\approx-1.45\). 3. b) Rewrite \(6^{2x-1}\) as \(\frac{36^x}{6}\). Then \(36^x=72\cdot4^x\). 4. Divide by \(4^x\): \(9^x=72\). Therefore, \(x=\frac{\log(72)}{\log(9)}\approx1.95\).

Answer

a) \(x\approx-1.45\) b) \(x\approx1.95\)

52841612

Solve each exponential equation. Round each result to two decimal places. a) \(3^x=2\cdot5^{x-1}\) b) \(7\cdot2^x=5\cdot3^x\) c) \(2^{x+1}\cdot5^x=0.8\cdot10^{2x}\)

Hints

- When \(x\) appears in exponents with different bases, collect the exponential terms into a ratio. - Combine powers with the same exponent before applying logarithms. - Simplify products such as \(2^x5^x\) using exponent properties. - Move all terms containing \(x\) to one side and constants to the other.

Solution

1. a) Rewrite the right side as \(\frac{2}{5}\cdot5^x\). Then \(\left(\frac{3}{5}\right)^x=\frac{2}{5}\), so \(x=\frac{\log(0.4)}{\log(0.6)}\approx1.79\). 2. b) Divide by \(5\cdot2^x\): \(\frac{7}{5}=\left(\frac{3}{2}\right)^x\). Thus, \(x=\frac{\log(1.4)}{\log(1.5)}\approx0.83\). 3. c) Simplify the left side: \(2^{x+1}\cdot5^x=2\cdot(2^x5^x)=2\cdot10^x\). 4. The equation becomes \(2\cdot10^x=0.8\cdot10^{2x}\). Divide by \(0.8\cdot10^x\): \(2.5=10^x\). Therefore, \(x=\log(2.5)\approx0.40\).

Answer

a) \(x=\frac{\log(0.4)}{\log(0.6)}\approx1.79\) b) \(x=\frac{\log(1.4)}{\log(1.5)}\approx0.83\) c) \(x=\log(2.5)\approx0.40\)

52842512

Ms. Miller invests \(\$2500.00\) at \(4\%\) annual interest. At the same time, Mr. Smith invests \(\$2800.00\) at \(0.8\%\) interest per quarter. After how many complete years will Ms. Miller's balance first be greater than Mr. Smith's balance?

Hints

- Write a compound-interest model for each investment. - Quarterly interest is applied four times per year. - Solve the equality to find the crossover time. - Check the complete years immediately before and after the crossover.

Solution

1. After \(n\) years, the balances are \(K_M(n)=2500\cdot1.04^n\) and \(K_S(n)=2800\cdot1.008^{4n}\). 2. Find the crossover time by solving \(2500\cdot1.04^n=2800\cdot1.008^{4n}\). 3. Rearranging gives \(\left(\frac{1.04}{1.008^4}\right)^n=1.12\), so \(n=\frac{\ln(1.12)}{\ln(1.04)-4\ln(1.008)}\approx15.42\). 4. At the end of year \(15\), Ms. Miller's balance is still smaller. At the end of year \(16\), her balance is larger. Therefore, the first complete year is \(16\).

Answer

After \(16\) complete years

52842612

Two savings plans are available. Option A starts with \(\$6000.00\) and earns \(2.5\%\) per year. Option B starts with \(\$5500.00\) and earns \(0.4\%\) per month. After how many complete months will Option B's balance first exceed Option A's balance?

Hints

- Use the same time unit for both plans. - Express Option A's annual growth over a number of months. - Solve the equality to find the crossover time. - Check the complete months immediately before and after the crossover.

Solution

1. Let \(m\) be the number of months. The balances are \(A(m)=6000\cdot1.025^{\frac{m}{12}}\) and \(B(m)=5500\cdot1.004^m\). 2. Find the crossover time by solving \(5500\cdot1.004^m=6000\cdot1.025^{\frac{m}{12}}\). 3. Taking logarithms gives \(m\left(\ln(1.004)-\frac{\ln(1.025)}{12}\right)=\ln\left(\frac{12}{11}\right)\). 4. Therefore, \(m\approx44.98\). At month \(44\), Option B is still lower; at month \(45\), it is higher. Thus, Option B first exceeds Option A after \(45\) complete months.

Answer

After \(45\) complete months

52848412

A biological population begins with \(250\) organisms and increases by \(12\%\) each hour. a) Write a function \(B(t)\) for the population after \(t\) hours. b) Find the modeled populations at \(t=1\), \(t=3.5\), \(t=8\), and \(t=12.5\). Round to the nearest whole organism. c) Determine when the population reaches \(400\), \(1000\), and \(2500\) organisms.

Hints

- Convert the percent increase to a growth factor. - Use the general exponential-growth form. - To solve for time, first isolate the exponential power. - Use logarithms to isolate the exponent.

Solution

1. The hourly growth factor is \(1.12\), so \(B(t)=250(1.12)^t\). 2. \(B(1)=280\), \(B(3.5)\approx371.71\approx372\), \(B(8)\approx618.99\approx619\), and \(B(12.5)\approx1030.78\approx1031\). 3. To reach a target \(y\), solve \(250(1.12)^t=y\), giving \(t=\frac{\ln(y/250)}{\ln(1.12)}\). 4. The population reaches \(400\) at about \(4.15\) hours, \(1000\) at about \(12.23\) hours, and \(2500\) at about \(20.32\) hours.

Answer

a) \(B(t)=250(1.12)^t\) b) \(280\), \(372\), \(619\), and \(1031\) organisms c) About \(4.15\), \(12.23\), and \(20.32\) hours, respectively

52850612

Solve each equation for \(x\). Round each answer to the nearest hundredth. a) \(5\cdot2^x=3^{x+1}\) b) \(8^{\frac{1}{x}}=2.4\) c) \(2\cdot4^x=3\cdot5^{x-1}\)

Hints

- When \(x\) appears in exponents on both sides, take logarithms and collect all terms containing \(x\). - Treat \(\frac{1}{x}\) as a factor after taking a logarithm. - Use exponent properties to simplify before applying logarithms. - Powers with the same exponent can be combined by dividing their bases.

Solution

1. a) Take natural logarithms: \(\ln(5)+x\ln(2)=(x+1)\ln(3)\). Collect the terms containing \(x\): \(x(\ln(2)-\ln(3))=\ln(3)-\ln(5)\). Thus, \(x=\frac{\ln(3)-\ln(5)}{\ln(2)-\ln(3)}\approx1.26\). 2. b) The exponent requires \(x\ne0\). Take natural logarithms: \(\frac{1}{x}\ln(8)=\ln(2.4)\). Therefore, \(x=\frac{\ln(8)}{\ln(2.4)}\approx2.38\), which satisfies the restriction. 3. c) Rewrite the equation as \(2\cdot4^x=\frac{3}{5}\cdot5^x\). Dividing by \(4^x\) and then by \(\frac{3}{5}\) gives \(\frac{10}{3}=\left(\frac{5}{4}\right)^x\). Therefore, \(x=\frac{\ln\left(\frac{10}{3}\right)}{\ln\left(\frac{5}{4}\right)}\approx5.40\).

Answer

a) \(x\approx1.26\) b) \(x\approx2.38\) c) \(x\approx5.40\)

52851412

Two bacterial cultures are compared. Culture A starts with \(1500\) bacteria and grows by \(15\%\) every \(4\) hours. Culture B starts with \(2500\) bacteria and grows by \(35\%\) every \(12\) hours. After how many hours will the cultures have the same population, and when will Culture A exceed Culture B?

Hints

- Write a growth model for each culture using its own growth interval. - Account for the different interval lengths in the exponents. - Set the populations equal to find the crossover time. - Use logarithms to isolate \(t\).

Solution

1. Model the populations by \(A(t)=1500\cdot1.15^{\frac{t}{4}}\) and \(B(t)=2500\cdot1.35^{\frac{t}{12}}\). 2. Set the models equal and take logarithms: \(\ln(1500)+\frac{t}{4}\ln(1.15)=\ln(2500)+\frac{t}{12}\ln(1.35)\). 3. Collect the terms containing \(t\): \(t\left(\frac{\ln(1.15)}{4}-\frac{\ln(1.35)}{12}\right)=\ln\left(\frac{2500}{1500}\right)\). 4. Therefore, \(t\approx51.43\) hours. 5. Culture A has the greater effective hourly growth rate, so it exceeds Culture B for \(t>51.43\) hours.

Answer

The populations are equal after about \(51.43\) hours. Culture A exceeds Culture B after that time.

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