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Precalculus math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Solve exponential equations

Learning goals

  • Rewrites expressions with a common base when possible.
  • Uses logarithms when common-base methods are not practical.
  • Finds the unknown exponent using equal bases or logarithms.
  • Checks solutions against domain and original equation.

Click problems to add them to your worksheet.

Solve exponential equations 1
Solve \(7^x=49\).

Hints

- Rewrite \(49\) as a power of \(7\). - Equal powers with the same positive base have equal exponents.

Solution

Since \(49=7^2\), the equation becomes \(7^x=7^2\). Therefore, \(x=2\).

Answer

\(x=2\)
Solve exponential equations 2
The graph shows \(y=2^x\) and the horizontal line \(y=6\). Use the graph to solve \(2^x=6\) to the nearest tenth.
Figure for problem 551489

Hints

- A solution is an x-value where the two graphs have the same y-value. - Locate the intersection of the exponential curve and the horizontal line. - Read the x-coordinate of that intersection to the requested precision.

Solution

1. A solution occurs where the two graphs intersect. 2. The intersection has x-coordinate about \(2.6\). 3. Therefore, \(2^x=6\) has solution \(x\approx2.6\) to the nearest tenth.

Answer

\(x\approx2.6\)
Solve exponential equations 3
Solve \(2^{x+1}=16\).

Hints

- Can you write the number on the right as a power of \(2\)? - When equal positive bases have equal powers, what must be true about their exponents?

Solution

1. Write \(16=2^4\), giving \(2^{x+1}=2^4\). 2. Set the exponents equal: \(x+1=4\), so \(x=3\).

Answer

\(x=3\)
Solve exponential equations 4
Solve \(9^x=3^8\) by writing both sides with base \(3\).

Hints

- Express the larger base as a power of the smaller base. - Apply the power-of-a-power rule before comparing exponents.

Solution

1. Since \(9=3^2\), the equation becomes \(3^{2x}=3^8\). 2. Thus \(2x=8\) and \(x=4\).

Answer

\(x=4\)
Solve exponential equations 5
Solve \(5^{-x}=\frac{1}{25}\).

Hints

- What does a negative exponent mean? - Write the fraction as a power with base \(5\).

Solution

1. Rewrite the right side as \(5^{-2}\). 2. Then \(-x=-2\), giving \(x=2\).

Answer

\(x=2\)
Solve exponential equations 6
Solve \(e^x=7\). Give an exact answer.

Hints

- Which function reverses raising \(e\) to a power? - How can you isolate the exponent without approximating it?

Solution

1. Apply the natural logarithm to both sides: \(\ln(e^x)=\ln(7)\). 2. Since \(\ln(e^x)=x\), the solution is \(x=\ln(7)\).

Answer

\(x=\ln(7)\)
Solve exponential equations 7
Solve \(3e^x=15\). Give an exact answer.

Hints

- First isolate the exponential expression. - Which operation undoes an exponential with base \(e\)?

Solution

1. Divide both sides by \(3\) to get \(e^x=5\). 2. Take natural logarithms to find \(x=\ln(5)\).

Answer

\(x=\ln(5)\)
Solve exponential equations 8
Does \(e^{x-1}+4=0\) have a real solution? Explain.

Hints

- Isolate the exponential expression. - What sign does \(e^t\) have for every real input \(t\)?

Solution

1. Subtracting \(4\) gives \(e^{x-1}=-4\). 2. Every real exponential value with base \(e\) is positive, so it cannot equal \(-4\). There is no real solution.

Answer

No real solution; \(e^{x-1}>0\) for all real \(x\).
Solve exponential equations 9
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) \(x=-\frac{1}{6}\log_2(3)\)
Solve exponential equations 10
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) \(x=2\)
Solve exponential equations 11
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) \(x=0\)
Solve exponential equations 12
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.
Solve exponential equations 13
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) \(\{-3,3\}\)
Solve exponential equations 14
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

\(x=2\ln(4)=\ln(16)\approx2.773\)
Solve exponential equations 15
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. \(135\,\text{cm}^2\)
Solve exponential equations 16
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

\(x\approx7.213\)
Solve exponential equations 17
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

\(x=1\)
Solve exponential equations 18
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

\(x=4\)
Solve exponential equations 19
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

\(x=3\)
Solve exponential equations 20
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

\(x=4\)
Solve exponential equations 21
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) \(x=-\frac{1}{2}\)
Solve exponential equations 22
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) \(x\approx6.58\)
Solve exponential equations 23
Solve \(8^x=4^{x+1}\) exactly.

Hints

- Can you express both bases as powers of a smaller common base? - Be careful to apply the exponent outside the parentheses to every term inside. - Compare the exponents after rewriting both sides.

Solution

1. Rewrite both bases as powers of \(2\): \(2^{3x}=2^{2(x+1)}\). 2. Equate exponents: \(3x=2x+2\). 3. Solving gives \(x=2\).

Answer

\(x=2\)
Solve exponential equations 24
Solve \(4^x-6\cdot2^x+8=0\). Start by setting \(u=2^x\).

Hints

- Express \(4^x\) in terms of \(2^x\). - Solve the resulting quadratic equation in the new variable. - Return to the original exponential expression for each candidate value.

Solution

1. Since \(4^x=(2^x)^2\), the equation becomes \(u^2-6u+8=0\). 2. Factoring gives \((u-2)(u-4)=0\), so \(u=2\) or \(u=4\). 3. Return to \(u=2^x\): \(2^x=2\) gives \(x=1\), and \(2^x=4\) gives \(x=2\).

Answer

\(x=1\) or \(x=2\).
Solve exponential equations 25
At what \(x\)-value do the graphed exponential curve \(f\) and horizontal line \(h\) intersect? Explain how an exponential equation gives the same answer.
Figure for problem 561590

Hints

- Read the height of the horizontal line from the graph. - Identify the base of the exponential curve, then write an equation for equal heights.

Solution

1. The horizontal line has height \(8\), so an intersection satisfies \(2^x=8\). 2. Since \(8=2^3\), we have \(x=3\), consistent with the graph.

Answer

\(x=3\)
Solve exponential equations 26
For each equation, choose the most useful first method: rewrite with a common base, take logarithms, or substitute \(u=2^x\). Write the first algebraic step, but do not solve. a) \(9^x=3^{x+2}\) b) \(5^x=7\) c) \(4^x-5\cdot2^x+4=0\)

Hints

- Decide whether the two sides can easily use the same exponential base. - Compare the number and arrangement of exponential terms in each equation before choosing a method. - In the last equation, look for a repeated expression that could simplify its structure.

Solution

1. For a), rewrite with base \(3\): \(3^{2x}=3^{x+2}\). 2. For b), take logarithms: \(x\ln(5)=\ln(7)\). 3. For c), set \(u=2^x\) and use \(4^x=u^2\), obtaining \(u^2-5u+4=0\).

Answer

a) Common base: \(3^{2x}=3^{x+2}\). b) Logarithms: \(x\ln(5)=\ln(7)\). c) Substitution: \(u^2-5u+4=0\), with \(u=2^x\).
Solve exponential equations 27
Find \(c\) so that the equation \(2^{x+c}=32\) has the solution \(x=2\).

Hints

- Put the required value of \(x\) into the equation. - Rewrite \(32\) as a power of the same base. - Solve for the exponent's missing constant.

Solution

1. Substitute the required solution: \(2^{2+c}=32=2^5\). 2. Thus \(2+c=5\) and \(c=3\).

Answer

\(c=3\)
Solve exponential equations 28
How many solutions does \(f(x)=h(x)\) have for the two graphed functions? Give the \(x\)-value of the solution and explain why there cannot be another.
Figure for problem 561593

Hints

- Compare whether the two curves rise or fall as you move to the right. - Read where the curves meet and write an equal-height equation to check the input. - Could a rising curve and a falling curve intersect again after they have crossed?

Solution

1. The increasing curve is \(f(x)=2^x\), and the decreasing curve is \(h(x)=4-2^x\). 2. Their intersection satisfies \(2^x=4-2^x\), so \(2\cdot2^x=4\) and \(2^x=2\). Hence \(x=1\). 3. As \(x\) increases, \(f\) increases while \(h\) decreases, so the two graphs can cross only once.

Answer

Exactly one solution, \(x=1\).
Solve exponential equations 29
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) \(x=1\)
Solve exponential equations 30
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) \(x=0\)
Solve exponential equations 31
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) \(x=-4\)
Solve exponential equations 32
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

\(\{-2,\ln(3),2\}\)
Solve exponential equations 33
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

\(\{2,3\}\)
Solve exponential equations 34
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) \(x=\frac{5}{2}\)
Solve exponential equations 35
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) \(x=-\frac{1}{2}\)
Solve exponential equations 36
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 \((\ln(3), -6)\)
Solve exponential equations 37
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

\(\{-2, 2\}\)
Solve exponential equations 38
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.
Solve exponential equations 39
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
Solve exponential equations 40
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) \(x=\frac{3}{10}\)
Solve exponential equations 41
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 \(x=\ln\!\left(\frac{e}{2}\right)=1-\ln(2)\).

Answer

\(x=1-\ln(2)\)
Solve exponential equations 42
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

\(x=\ln(4)\)
Solve exponential equations 43
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

\(\left\{-\sqrt{\ln(8)},-\sqrt{\ln(4)},\sqrt{\ln(4)},\sqrt{\ln(8)}\right\}\)
Solve exponential equations 44
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

\(x=2\ln(2)=\ln(4)\approx1.386\)
Solve exponential equations 45
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. \(A(t)=150e^{0.0497t}\)
Solve exponential equations 46
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 \(D(t)=72.0e^{-0.0286t}\)
Solve exponential equations 47
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
Solve exponential equations 48
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? Round to the nearest hundredth of a centimeter. b) Find when the tree reaches a height of \(2.5\,\text{m}\). Round the time to the nearest hundredth of a year, and verify using the exact time.

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 exact 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, \(h(4)=\frac{3.2e^2}{e^2+7}\approx1.6432613\,\text{m}\). 3. The growth is \(100(h(4)-h(0))\approx124.33\,\text{cm}\). 4. Set \(h(t)=2.5\): \(2.5(e^{0.5t}+7)=3.2e^{0.5t}\). Thus, \(0.7e^{0.5t}=17.5\) and \(e^{0.5t}=25\). 5. The exact time is \(t=2\ln(25)\approx6.44\) years. Substitution gives \(h(2\ln(25))=\frac{3.2(25)}{25+7}=2.5\,\text{m}\).

Answer

a) Approximately \(124.33\,\text{cm}\) b) \(t=2\ln(25)\approx6.44\) years; substituting the exact time gives \(h(t)=2.5\,\text{m}\).
Solve exponential equations 49
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) \(\{\ln(3)\}\)
Solve exponential equations 50
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) \(\{5\}\)
Solve exponential equations 51
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
Solve exponential equations 52
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 \(x=4\)
Solve exponential equations 53
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: \(x=\ln(3)\)
Solve exponential equations 54
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 \(200\,{}^\circ\text{C}\). Round the time to the nearest hundredth of a minute.

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. Use \(T(t)=S-(S-T_0)e^{-kt}\). With \(S=220\) and \(T_0=20\), this gives \(T(t)=220-200e^{-kt}\). 2. Use \(T(15)=120\): \(120=220-200e^{-15k}\), so \(e^{-15k}=\frac{1}{2}\) and \(k=\frac{\ln(2)}{15}\). 3. The exact model is \(T(t)=220-200e^{-(\ln(2)/15)t}\). 4. Set \(T(t)=200\): \(e^{-kt}=0.1\). Thus, \(t=\frac{\ln(10)}{k}=\frac{15\ln(10)}{\ln(2)}\approx49.83\) minutes.

Answer

a) \(T(t)=220-200e^{-(\ln(2)/15)t}\) b) \(t=\frac{15\ln(10)}{\ln(2)}\approx49.83\,\text{minutes}\)
Solve exponential equations 55
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

\(x=\frac{\ln(3)}{\ln(4)}\approx0.79\)
Solve exponential equations 56
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

\(x=\frac{\ln(8)}{\ln(3)}\approx1.89\)
Solve exponential equations 57
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 \(13.29\,\text{minutes}\)
Solve exponential equations 58
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 \(6.64\,\text{hours}\)
Solve exponential equations 59
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.
Solve exponential equations 60
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) \(x\in\{-3, 3\}\)
Solve exponential equations 61
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) \(\{-2, 1\}\)
Solve exponential equations 62
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

\(\log(5) \approx 0.70\)
Solve exponential equations 63
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

\(\log_2(3) \approx 1.58\)
Solve exponential equations 64
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

\(x\approx-1.10\)
Solve exponential equations 65
Both accounts earn interest once a year, with no deposits or withdrawals. For each account, find the first whole year when its balance reaches or exceeds the target. 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

- Which yearly growth factor applies in each account? - Find when each power first reaches the target, then think about whole-year interest payments. - If the equation gives part of a year, which whole year is first large enough?

Solution

1. In part a, \(15{,}000=10{,}000\cdot1.04^n\), so \(1.5=1.04^n\). 2. Taking logarithms gives \(n=\frac{\log(1.5)}{\log(1.04)}\approx10.34\). Since interest is added at the end of each whole year, the first qualifying year is \(11\). 3. In part b, \(1200=500\cdot1.06^n\), so \(2.4=1.06^n\). 4. Taking logarithms gives \(n=\frac{\log(2.4)}{\log(1.06)}\approx15.02\). The first qualifying whole year is \(16\).

Answer

a) \(11\) whole years b) \(16\) whole years
Solve exponential equations 66
A rare bird population in a wildlife preserve is modeled as growing by \(6\%\) per year. At the start of the study, there are \(80\) birds. Assume the exponential model also applies to fractional years. a) Find the predicted population after \(15\) years. Round to the nearest whole bird. b) After how many years will the modeled population have tripled? Round to the nearest tenth of a year.

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. After \(15\) years, \(B(15)=80\cdot1.06^{15}\approx191.72\), so the model predicts \(192\) birds. 3. For a tripled population, set \(80\cdot1.06^t=3\cdot80\), giving \(1.06^t=3\). 4. Thus, \(t=\frac{\ln(3)}{\ln(1.06)}\approx18.9\) years.

Answer

a) About \(192\) birds b) About \(18.9\) years
Solve exponential equations 67
Two cities are growing at different rates. Northgate has \(25{,}000\) residents and grows by \(3.5\%\) per year. Lakehaven 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? Round to the nearest tenth of a year.

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: \(t=\frac{\ln(1.6)}{\ln(1.035)-\ln(1.012)}\approx20.91\), which rounds to \(20.9\) years.

Answer

About \(20.9\) years
Solve exponential equations 68
Let \(h(x)=6^x\). a) Use logarithms to find the amount \(k\) by which the input must increase so that the function value is multiplied by \(100\). Round to three decimal places. b) Determine how this value of \(k\) changes for \(g(x)=36^x\). Justify your answer using algebra or exponent properties.

Hints

- Set up an equation using \(h(x+k)=100h(x)\). - Solve an equation of the form \(a^k=b\) with logarithms. - Express \(36\) as a power of \(6\). - Use the power-of-a-power property to compare the two equations.

Solution

1. a) Set \(h(x+k)=100h(x)\): \(6^{x+k}=100\cdot6^x\). 2. Divide by \(6^x\): \(6^k=100\). Therefore, \(k=\log_6(100)=\frac{\log(100)}{\log(6)}\approx2.570\). 3. b) For \(g\), the equation is \(36^k=100\). Since \(36=6^2\), \(6^{2k}=100\). 4. Compared with \(6^{k_{\!6}}=100\), this gives \(2k_{\!36}=k_{\!6}\). Thus, \(k_{\!36}=\frac{1}{2}k_{\!6}\approx1.285\).

Answer

a) \(k\approx2.570\) b) The required increase is halved: \(k\approx1.285\).
Solve exponential equations 69
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) \(x\approx1.08\)
Solve exponential equations 70
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

\((3, 1458)\)
Solve exponential equations 71
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

\((2.26, 15.10)\)
Solve exponential equations 72
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) \(x=\frac{\log(3)}{\log(2)}\approx1.58\)
Solve exponential equations 73
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
Solve exponential equations 74
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 \(x=-2\)
Solve exponential equations 75
A yeast culture starts with \(250\) cells and grows by \(12\%\) each hour. Use the exponential model for parts of an hour too. a) Write a function \(B(t)\) for the number of cells after \(t\) hours. b) Find the modeled cell counts at \(t=1\), \(t=3.5\), \(t=8\), and \(t=12.5\). Round to the nearest whole cell. c) Find when the model reaches \(400\), \(1000\), and \(2500\) cells. Round each time to the nearest hundredth of an hour.

Hints

- What factor represents a \(12\%\) increase over one hour? - How do you find the amount after part of an hour in this model? - What equation would give the time for a chosen population size? - Keep the full calculator value until the final rounding step.

Solution

1. The hourly growth factor is \(1.12\), so \(B(t)=250(1.12)^t\). 2. \(B(1)=280\), \(B(3.5)\approx371.71\), \(B(8)\approx618.99\), and \(B(12.5)\approx1030.78\). To the nearest whole cell, these are \(280\), \(372\), \(619\), and \(1031\). 3. For any target \(y\), solve \(250(1.12)^t=y\), giving \(t=\frac{\ln(y/250)}{\ln(1.12)}\). 4. The times for \(y=400\), \(1000\), and \(2500\) are approximately \(4.15\), \(12.23\), and \(20.32\) hours.

Answer

a) \(B(t)=250(1.12)^t\) b) \(280\), \(372\), \(619\), and \(1031\) cells c) Approximately \(4.15\), \(12.23\), and \(20.32\) hours, respectively
Solve exponential equations 76
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 \(x=-2\)
Solve exponential equations 77
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) \(x\approx5.70\)
Solve exponential equations 78
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

\(x=\frac{\log(6)}{\log(6)-\log(5)}\approx9.83\)
Solve exponential equations 79
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

\(x=\log_3(6)\approx1.631\)
Solve exponential equations 80
Rosa's investment starts at \(\$12{,}500.00\) and is modeled to grow by \(6.5\%\) per year. Minh's investment starts at \(\$20{,}000.00\) and grows by \(3.8\%\) per year in the model. Use both exponential models for parts of a year too. When will the balances be equal, and after what time will Rosa's exceed Minh's? Round the equality time to the nearest tenth of a year.

Hints

- Write one exponential model for each starting balance and growth rate. - What equation gives the time when the balances agree? - Which investment grows faster once their balances are equal?

Solution

1. Model the balances by \(R(t)=12{,}500\cdot1.065^t\) and \(M(t)=20{,}000\cdot1.038^t\). 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, \(t=\frac{\ln(1.6)}{\ln\left(\frac{1.065}{1.038}\right)}\approx18.30\), or \(18.3\) years to the nearest tenth. 5. Since Rosa's investment has the larger growth factor, her balance exceeds Minh's when \(t>\frac{\ln(1.6)}{\ln\left(\frac{1.065}{1.038}\right)}\).

Answer

Equal after about \(18.3\) years; Rosa's balance is larger when \(t>\frac{\ln(1.6)}{\ln\left(\frac{1.065}{1.038}\right)}\), where \(t\) is in years.
Solve exponential equations 81
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
Solve exponential equations 82
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

\(2035\)
Solve exponential equations 83
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
Solve exponential equations 84
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
Solve exponential equations 85
Samira drinks a cup of coffee containing \(120\,\text{mg}\) of caffeine at 2:00 p.m. Suppose the amount in her body decreases by \(13\%\) each hour. Use an exponential model for parts of an hour too. 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. Round to the nearest hundredth of a milligram. c) Suppose this person uses \(15\,\text{mg}\) as a personal planning cutoff. Find approximately what clock time the modeled amount reaches this level, to the nearest minute. After that time, the modeled amount is below the cutoff.

Hints

- What fraction of the caffeine remains after each hour? - How many hours pass between the two times in part b)? - What equation gives the time when the model reaches the personal cutoff? - How do you convert a decimal number of hours into 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, so \(K(8)=120(0.87)^8\approx39.39\,\text{mg}\). 3. Set \(120(0.87)^t=15\). Then \(0.87^t=0.125\), so \(t=\frac{\ln(0.125)}{\ln(0.87)}\approx14.93186\) hours. 4. This is about \(14\) hours \(56\) minutes after 2:00 p.m., or about 4:56 a.m. the next day. The modeled amount falls below the cutoff after the crossover.

Answer

a) \(K(t)=120(0.87)^t\) b) About \(39.39\,\text{mg}\) c) About 4:56 a.m. the next day; below \(15\,\text{mg}\) afterward.
Solve exponential equations 86
The area of a park pond covered by water lilies increases by \(40\%\) each week. At the start of the observation (\(t=0\)), the lilies cover \(2\,\text{m}^2\). Use the same exponential model for parts of a week and for earlier times. 1. Write a function \(A(t)\) that gives the covered area after \(t\) weeks. 2. Find the modeled covered area after \(4\) weeks and \(2\) weeks before the observation began. Round both to the nearest hundredth of a square meter. 3. Find when the modeled area reaches \(20\,\text{m}^2\). Round to the nearest hundredth of a week.

Hints

- How does a percentage increase become a weekly growth factor? - What does a negative input mean when \(t=0\) is the start of observation? - How can you solve for the time when the area reaches a given value? - Keep full calculator precision until the final answer.

Solution

1. A \(40\%\) weekly increase gives the growth factor \(1.4\). Thus, \(A(t)=2(1.4)^t\). 2. \(A(4)=2(1.4)^4=7.6832\approx7.68\,\text{m}^2\). For two weeks before the start, use \(t=-2\): \(A(-2)=2(1.4)^{-2}\approx1.02\,\text{m}^2\). 3. Solve \(2(1.4)^t=20\), or \(1.4^t=10\). Then \(t=\frac{\ln(10)}{\ln(1.4)}\approx6.84\) weeks.

Answer

1. \(A(t)=2(1.4)^t\) 2. \(7.68\,\text{m}^2\) after four weeks and \(1.02\,\text{m}^2\) two weeks earlier 3. About \(6.84\) weeks
Solve exponential equations 87
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 modeled number of bacteria after \(8\) hours. Round to the nearest whole bacterium. c) Find when the modeled number of bacteria is five times the initial number. Round to the nearest hundredth of an hour.

Hints

- Which table entry gives the value of \(c\)? - Compare outputs one hour apart to find the growth factor. - How do you convert an output in thousands to a count of bacteria? - Use the factor of increase to set up an equation for the time.

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. 2. \(f(8)=400(1.3)^8\approx3262.922884\) thousand. This is approximately \(3{,}262{,}923\) bacteria, rounded to a whole bacterium. 3. A fivefold increase satisfies \(1.3^x=5\). Therefore, \(x=\frac{\ln(5)}{\ln(1.3)}\approx6.13\) hours.

Answer

a) \(f(x)=400(1.3)^x\), where \(f(x)\) is in thousands b) About \(3{,}262{,}923\) bacteria c) About \(6.13\) hours
Solve exponential equations 88
A bacterial culture grows exponentially. At \(t=0\), the culture contains \(2500\) bacteria. After \(6\) hours, the population has tripled. Use the model for parts of an hour as well. a) Find a model of the form \(N(t)=N_0b^t\). b) Find the modeled number of bacteria after \(15\) hours. Round to the nearest whole bacterium. c) Find when the model reaches \(1{,}000{,}000\) bacteria. Round to the nearest hundredth of an hour.

Hints

- How is the six-hour tripling factor related to the one-hour base \(b\)? - In part b), how many six-hour periods does the time represent? - Which population ratio appears when you set the model equal to the target?

Solution

1. Since the population triples in \(6\) hours, \(b^6=3\), so \(b=3^{1/6}\). Thus, \(N(t)=2500(3^{1/6})^t=2500\cdot3^{t/6}\). 2. \(N(15)=2500\cdot3^{15/6}\approx38{,}971.14\), or about \(38{,}971\) bacteria. 3. Solve \(1{,}000{,}000=2500\cdot3^{t/6}\). Then \(3^{t/6}=400\), so \(t=\frac{6\ln(400)}{\ln(3)}\approx32.72\) hours.

Answer

a) \(N(t)=2500(3^{1/6})^t\), or \(N(t)=2500\cdot3^{t/6}\) b) About \(38{,}971\) bacteria c) About \(32.72\) hours
Solve exponential equations 89
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. Round to the nearest cent. c) Find when the value decreases to one-fourth of the original purchase price. Round to the nearest hundredth of a year.

Hints

- What does the coefficient tell you about the value when \(t=0\)? - How is the yearly multiplier related to the percent lost each year? - For one-fourth of the original value, which part of the formula needs to equal \(0.25\)?

Solution

1. At \(t=0\), \(W(0)=3800\), so the original price was \(\$3800.00\). The factor \(0.84\) means the bike loses \(16\%\) of its value each year. 2. \(W(4)=3800(0.84)^4\approx1891.91\), so the value is about \(\$1891.91\). 3. One-fourth of the initial value satisfies \(0.84^t=0.25\). Thus, \(t=\frac{\ln(0.25)}{\ln(0.84)}\approx7.95\) years.

Answer

a) Original price: \(\$3800.00\); annual decrease: \(16\%\) b) About \(\$1891.91\) c) About \(7.95\) years
Solve exponential equations 90
Solve \(7\cdot3^{2x-1}=4\cdot2^{x+3}\). Give an exact answer and a decimal approximation to the nearest ten-thousandth.

Hints

- What operation can you use to bring the unknown exponents down as factors? - Which logarithm rules apply to the numerical factors and powers? - Collect and factor the terms containing \(x\).

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

\(x=\frac{5\ln(2)+\ln(3)-\ln(7)}{2\ln(3)-\ln(2)}\approx1.7409\)
Solve exponential equations 91
Solve \(2^{x+1}5^{2x}=10^{x-1}5^4\). Give an exact answer and a decimal approximation to the nearest ten-thousandth.

Hints

- Can you write base \(10\) as a product of bases that already appear? - How does dividing like bases simplify the exponent expressions? - After simplifying, how can you isolate the unknown exponent?

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

\(x=3-\frac{\ln(4)}{\ln(5)}\approx2.1386\)
Solve exponential equations 92
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 intensity at a depth of \(10\,\text{m}\). Round to the nearest hundredth of a watt per square meter. b) Find the depth at which the intensity is \(50\,\text{W}/\text{m}^2\). Round to the nearest hundredth of a meter. c) The model can be written as \(I(d)=I_0e^{-kd}\). Find \(k\) in inverse meters, rounded to four decimal places.

Hints

- How does the intensity change over each additional three meters? - How many three-meter intervals correspond to a depth of \(d\)? - What ratio of initial intensity remains at the target depth? - Compare the three-meter reduction factors 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. \(I(10)=400(0.82)^{10/3}\approx206.43\,\text{W}/\text{m}^2\). 3. For \(I(d)=50\), solve \(0.82^{d/3}=0.125\). Thus, \(d=\frac{3\ln(0.125)}{\ln(0.82)}\approx31.44\,\text{m}\). 4. Since \(e^{-3k}=0.82\), \(k=-\frac{\ln(0.82)}{3}\approx0.0662\,\text{m}^{-1}\).

Answer

a) About \(206.43\,\text{W}/\text{m}^2\) b) About \(31.44\,\text{m}\) c) \(k\approx0.0662\,\text{m}^{-1}\)
Solve exponential equations 93
A bacterial culture is treated with an antibiotic. Suppose its population follows an exponential model that decreases by \(15\%\) every \(20\) minutes, including for parts of a twenty-minute interval. At \(t=0\), there are \(500{,}000\) bacteria. a) How many bacteria remain after exactly \(2\) hours? Round to the nearest whole bacterium. b) Find when the population has decreased to \(1\%\) of its initial value. Round to the nearest hundredth of a minute. c) Find the percent decrease per minute. Round to the nearest hundredth of a percent.

Hints

- How many twenty-minute intervals are there in two hours? - What remaining fraction corresponds to a \(1\%\) target? - How can you find the growth factor for one minute from the factor for twenty minutes? - Convert the one-minute factor to a percentage 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 \(N(120)=500{,}000(0.85)^6\approx188{,}575\) bacteria. 3. For \(1\%\) remaining, solve \(0.85^{t/20}=0.01\). Thus, \(t=\frac{20\ln(0.01)}{\ln(0.85)}\approx566.72\) minutes. 4. The one-minute factor is \(0.85^{1/20}\approx0.991907\). Thus, the decrease is about \(0.81\%\) per minute.

Answer

a) About \(188{,}575\) bacteria b) About \(566.72\) minutes c) About \(0.81\%\) per minute
Solve exponential equations 94
The fictional city of Pine Ridge had \(450{,}000\) residents in 2005 and \(585{,}000\) in 2020. Assume an exponential growth model, with time measured in years after the start of 2005. a) Find the average annual growth rate, rounded to the nearest hundredth of a percent. b) Predict the population in 2045 if the same rate continues. Round to the nearest hundred people. c) Find the calendar year during which the modeled population first exceeds one million.

Hints

- How many years separate the two measured populations? - Find the yearly factor that produces the observed overall growth. - Convert that factor to an annual percentage. - How does an elapsed time convert to a calendar year?

Solution

1. Fifteen years elapse from 2005 to 2020. Let \(P(t)=450{,}000q^t\). Then \(585{,}000=450{,}000q^{15}\), giving \(q=1.3^{1/15}\approx1.0176448\) and a growth rate of about \(1.76\%\) per year. 2. For 2045, \(t=40\): \(P(40)=450{,}000q^{40}\approx905{,}861\), or about \(905{,}900\) people to the nearest hundred. 3. Set \(450{,}000q^t=1{,}000{,}000\). Then \(t=\frac{\ln(1{,}000{,}000/450{,}000)}{\ln(q)}\approx45.65\). The model passes the target during 2050.

Answer

a) About \(1.76\%\) per year b) About \(905{,}900\) people c) During 2050
Solve exponential equations 95
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) Using the rounded value of \(k\) from part a), find the temperature after \(15\) minutes. Round to the nearest tenth of a degree Celsius.

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. - For part c), use the rounded constant requested in part a).

Solution

1. Use \(T(10)=44\): \(20+60e^{-10k}=44\). Then \(e^{-10k}=0.4\), so \(k=-\frac{\ln(0.4)}{10}\approx0.092\). 2. For \(T(t)=25\), \(20+60e^{-kt}=25\), so \(e^{-kt}=\frac{1}{12}\). Using the exact expression for \(k\), \(t=\frac{10\ln(12)}{-\ln(0.4)}\approx27.1\) minutes. 3. Using \(k=0.092\), \(T(15)=20+60e^{-0.092(15)}\approx35.1\,{}^\circ\text{C}\).

Answer

a) \(k\approx0.092\) b) About \(27.1\) minutes c) About \(35.1\,{}^\circ\text{C}\)
Solve exponential equations 96
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
Solve exponential equations 97
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. Round to the nearest hundredth of a year.
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. - Translate “one-fourth of its original price” into a target expressed in terms of the initial value.

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 the initial value gives the equation \(800(0.75)^t=200\). Thus, \(t=\frac{\ln(0.25)}{\ln(0.75)}\approx4.82\) years.

Answer

a) Initial value: \(\$800\); factor: \(0.75\); \(W(t)=800(0.75)^t\) b) \(25\%\) decrease per year c) About \(4.82\) years
Solve exponential equations 98
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 \(9\,\text{m}\). Round to the nearest hundredth of a year. c) Identify the theoretical height the tree approaches but does not exceed. Justify the value from the function.

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 \(t=\frac{\ln3}{0.15}\approx7.32\) 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) \(12\,\text{m}\)
Solve exponential equations 99
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 \(B(4)=40\). Thus, \(40=80e^{4k_B}\), so \(k_B=-\frac{\ln 2}{4}\). Therefore, \(B(t)=80e^{-(\ln 2/4)t}\). 3. Set the models equal: \(20e^{(\ln 2/2)t}=80e^{-(\ln 2/4)t}\). This simplifies to \(e^{(3\ln 2/4)t}=4\). Taking natural logarithms gives \(\frac{3\ln 2}{4}t=2\ln 2\), so \(t=\frac{8}{3}\).

Answer

a) \(A(t)=20e^{(\ln 2/2)t}\) and \(B(t)=80e^{-(\ln 2/4)t}\) b) \(t=\frac{8}{3}\)
Solve exponential equations 100
A bacterial culture grows exponentially. a) Find a model of the form \(f(t)=ce^{kt}\). b) Find when the culture reaches \(1000\) bacteria. Round to the nearest hundredth of an hour.
Figure for problem 534491

Hints

- What initial count can you read where the curve meets the y-axis? - Use another marked point to find the growth factor over a fixed time. - What population ratio must the function reach in part b)?

Solution

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

Answer

a) \(f(t)=100e^{(\ln(2)/10)t}\) b) About \(33.22\) hours
Solve exponential equations 101
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 \(W(t)=500\left(1-e^{-0.3466t}\right)\)
Solve exponential equations 102
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) \(t<4\)
Solve exponential equations 103
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

\(x=2\)
Solve exponential equations 104
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) \(T_{\min}=10\ln(1.2)\approx 1.82\,\text{weeks}\)
Solve exponential equations 105
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

\(x\approx6.66\)
Solve exponential equations 106
Solve the inequality without using logarithms. Give the solution in interval notation. \((0.5)^{x^2-1}\le4^{x-1}\)

Hints

- Rewrite both sides as powers with base \(2\). - Express \(0.5\) as the reciprocal of \(2\). - Because the common base is greater than \(1\), the inequality direction stays the same when comparing exponents. - Solve the resulting quadratic inequality by locating the zeros and testing intervals.

Solution

1. Rewrite both sides with base \(2\): \((2^{-1})^{x^2-1}\le(2^2)^{x-1}\). 2. Simplify the exponents: \(2^{-x^2+1}\le2^{2x-2}\). 3. Since the base \(2>1\), compare the exponents without reversing the inequality: \(-x^2+1\le2x-2\). 4. Rearrange: \(x^2+2x-3\ge0\). 5. Factor: \((x+3)(x-1)\ge0\). The product is nonnegative outside the interval between its zeros. 6. Therefore, \(x\le-3\) or \(x\ge1\).

Answer

\((-\infty, -3] \cup [1, \infty)\)
Solve exponential equations 107
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 \(9\)
Solve exponential equations 108
Pinecrest Timber has \(800\,\text{m}^3\) of timber and grows its inventory by \(4\%\) per year. Blue River Forestry has \(2000\,\text{m}^3\) and grows its inventory by \(1.5\%\) per year. After how many years will Pinecrest Timber's inventory be exactly \(75\%\) of Blue River Forestry's inventory? Round to the nearest tenth of a year.

Hints

- Write an exponential growth model for each inventory. - Express \(75\%\) of the second inventory as multiplication by \(0.75\). - Set the first company'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: \(t=\frac{\ln(1.875)}{\ln(1.04)-\ln(1.015)}\approx25.83\), which rounds to \(25.8\) years.

Answer

About \(25.8\) years
Solve exponential equations 109
Hot cocoa is placed in a room at a constant temperature of \(20\,{}^\circ\text{C}\). The cocoa begins at \(80\,{}^\circ\text{C}\). 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, rounded to two decimal places. c) Find the cocoa's actual temperature after \(10\) minutes, rounded to two decimal places. d) When does the temperature difference first become less than \(5\,{}^\circ\text{C}\)? Give the boundary time to two decimal places and the first whole-number minute after the boundary.

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 \(80-20=60\,{}^\circ\text{C}\), and the decay factor is \(0.88\). Thus, \(f(x)=60(0.88)^x\). 2. The differences are \(f(5)\approx31.66\,{}^\circ\text{C}\) and \(f(15)\approx8.82\,{}^\circ\text{C}\). 3. After \(10\) minutes, the difference is \(f(10)\approx16.71\,{}^\circ\text{C}\). Add room temperature to get \(36.71\,{}^\circ\text{C}\). 4. From \(60(0.88)^x<5\), the boundary is \(x=\frac{\ln(5/60)}{\ln(0.88)}\approx19.44\). The temperature difference is below the threshold after this time, first at whole minute \(20\).

Answer

a) \(f(x)=60(0.88)^x\) b) After \(5\) minutes: \(31.66\,{}^\circ\text{C}\); after \(15\) minutes: \(8.82\,{}^\circ\text{C}\) c) \(36.71\,{}^\circ\text{C}\) d) Boundary time approximately \(19.44\) minutes; first whole-number minute below the threshold: \(20\)
Solve exponential equations 110
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. Round nonwhole amounts to the nearest hundredth of a milligram. b) When does the modeled amount reach \(100\,\text{mg}\)? Round to the nearest hundredth of an hour. c) When does the modeled amount reach one-tenth of the initial amount? Round to the nearest hundredth of an hour.

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 starting amount is \(40\,\text{mg}\). Solve \(400(0.85)^t=40\), obtaining \(t=\frac{\ln(0.1)}{\ln(0.85)}\approx14.17\) 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
Solve exponential equations 111
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) \(x\approx1.95\)
Solve exponential equations 112
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) \(x=\log(2.5)\approx0.40\)
Solve exponential equations 113
Ms. Miller invests \(\$2500.00\) at \(4\%\) annual interest, compounded once a year. At the same time, Mr. Smith invests \(\$2800.00\) at \(0.8\%\) interest per quarter, compounded every quarter. After how many complete years will Ms. Miller's balance first be greater than Mr. Smith's balance?

Hints

- How often is interest added to each balance? - Write both balances in terms of the same number of years. - When do the two models give equal balances? - Which whole year first comes 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. At the crossover, \(2500\cdot1.04^n=2800\cdot1.008^{4n}\), so \(\left(\frac{1.04}{1.008^4}\right)^n=1.12\). 3. Taking logarithms gives \(n=\frac{\ln(1.12)}{\ln(1.04)-4\ln(1.008)}\approx15.42\). 4. At the end of year \(15\), Ms. Miller has about \(\$4502.36\), less than Mr. Smith's \(\$4516.37\). At the end of year \(16\), she has about \(\$4682.45\), more than Mr. Smith's \(\$4662.64\).

Answer

After \(16\) complete years.
Solve exponential equations 114
Nina has \(\$6000.00\) in a savings account with an effective annual growth rate of \(2.5\%\). Mateo has \(\$5500.00\) in an account that grows by \(0.4\%\) each month. Use the exponential model for Nina's account between whole years too. After how many complete months will Mateo's balance first exceed Nina's?

Hints

- How can you express a yearly growth rate using months as the time unit? - Write a balance model for each plan. - Find where the two balances would be equal, then check whole months around that time.

Solution

1. Let \(m\) be the number of months. Nina's balance is \(N(m)=6000\cdot1.025^{m/12}\), and Mateo's is \(M(m)=5500\cdot1.004^m\). 2. To find when the balances match, solve \(5500\cdot1.004^m=6000\cdot1.025^{m/12}\). 3. Taking logarithms gives \(m\left(\ln(1.004)-\frac{\ln(1.025)}{12}\right)=\ln\left(\frac{12}{11}\right)\), so \(m\approx44.98\). 4. Mateo's balance is lower at month \(44\) but higher at month \(45\), so the answer is \(45\) complete months.

Answer

After \(45\) complete months.
Solve exponential equations 115
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) \(x\approx5.40\)
Solve exponential equations 116
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. Use exponential models for any part of those growth intervals. After how many hours will the modeled populations be equal, and when will Culture A exceed Culture B? Round the equality time to the nearest hundredth of an hour.

Hints

- Use the number of four-hour or twelve-hour intervals in each exponent. - How can the two population models be compared at a common time? - What determines which culture grows faster in the long run? - Keep the crossover value unrounded until the final answer.

Solution

1. Model the populations by \(A(t)=1500\cdot1.15^{t/4}\) and \(B(t)=2500\cdot1.35^{t/12}\). 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. Culture A has the greater effective hourly growth rate, so its modeled population is larger for times after the crossover.

Answer

Equal after about \(51.43\) hours; Culture A exceeds Culture B after that time.

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