- Identify the base, exponent, and exponential result. - A logarithm asks which exponent on the base produces the argument.

1. In \(b^c=a\), the equivalent logarithmic statement is \(\log_b(a)=c\). 2. Therefore, \(4^3=64\) becomes \(\log_4(64)=3\).

\(\log_4(64)=3\)

- Read the point as an input and its corresponding output. - What happens to these coordinates when the operation is reversed?

1. Inverse functions reverse each input-output pair. 2. The point \((2, 9)\) on \(f\) gives \((9, 2)\) on \(f^{-1}\).

\((9, 2)\)

- In \(b^c=a\), which number is the base? - A logarithm asks for the exponent, not the result of exponentiation.

1. The logarithm gives the exponent needed to get \(8\) from base \(2\). 2. Therefore, \(\log_{2}(8)=3\).

\(\log_{2}(8)=3\)

- Identify the base and the exponent in the logarithm statement. - Write the number inside the logarithm as the output of a power.

1. The equation says that base \(3\) raised to the fourth power gives \(81\). 2. Thus, \(3^4=81\).

\(3^4=81\)

- Find the requested number in the output row. - What input produced that output?

1. The output \(4\) occurs when the input is \(2\). 2. Since a logarithm asks for the exponent that gives the output, \(\log_{2}(4)=2\).

\(2\)

- What sign can an output of \(2^x\) have? - A logarithm reverses an exponential; can that exponential produce the proposed input?

1. The exponential \(2^x\) is positive for every real \(x\). 2. No real exponent produces \(-4\), so \(\log_{2}(-4)\) is not real.

No. \(2^x>0\) for every real \(x\).

- Ask what power of the base produces \(2\). - How do negative exponents change a fraction?

1. Since \(\left(\frac{1}{2}\right)^{-1}=2\), the required exponent is \(-1\). 2. Therefore, \(\log_{\frac{1}{2}}(2)=-1\).

\(-1\)

- Rewrite a logarithmic statement as an exponential statement. - Ask which exponent on the base produces the argument. - Express fractions and decimals as powers with negative exponents. - Isolate the logarithm before solving a logarithmic equation.

1. a) Since \(2^7 = 128\), \(\log_2(128) = 7\). 2. b) Since \(3^{-2} = \frac{1}{9}\), \(\log_3\left(\frac{1}{9}\right) = -2\). 3. c) Since \(10^{-1} = 0.1\), \(\log_{10}(0.1) = -1\). Therefore, \(4 \cdot (-1) = -4\). 4. d) Rewrite the equation as \(x = 5^2\), so \(x = 25\). 5. e) Divide by \(3\) to get \(\log_2(x) = 4\). Then \(x = 2^4 = 16\).

a) \(7\) b) \(-2\) c) \(-4\) d) \(x = 25\) e) \(x = 16\)

- Ask which exponent on the base produces the argument. - Rewrite reciprocals as powers with negative exponents. - Express a square root as a rational exponent. - What exponent on a valid base produces \(1\)?

1. a) Since \(6^3 = 216\), \(\log_6(216) = 3\). 2. b) Since \(5^0 = 1\), \(\log_5(1) = 0\). 3. c) Rewrite \(0.25\) as \(\frac{1}{4} = 2^{-2}\). Therefore, \(\log_2(0.25) = -2\). 4. d) Since \(\sqrt{10} = 10^{1/2}\), \(\log_{10}(\sqrt{10}) = \frac{1}{2}\).

a) \(3\) b) \(0\) c) \(-2\) d) \(\frac{1}{2}\)

- Rewrite each argument as a power of its logarithm base. - Converting a decimal to a fraction may reveal a familiar power. - Relate roots to rational exponents. - A reciprocal corresponds to a negative exponent.

1. a) Since \(2^7 = 128\), \(\log_2(128) = 7\). 2. b) Since \(3 = \sqrt{9} = 9^{1/2}\), \(\log_9(3) = \frac{1}{2}\). 3. c) Rewrite \(0.008\) as \(\frac{1}{125} = 5^{-3}\). Therefore, \(\log_5(0.008) = -3\). 4. d) Rewrite \(\frac{1}{\sqrt{10}}\) as \(10^{-1/2}\). Therefore, \(\log_{10}\left(\frac{1}{\sqrt{10}}\right) = -\frac{1}{2}\).

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

- What does a logarithm tell you about the exponent in an exponential equation? - Identify the base, exponent, and power in each equation. - Write \(0.001\) as a power of \(10\). - Recall how a negative exponent produces a reciprocal.

1. Use \(b^c = a\) if and only if \(\log_b(a) = c\). 2. a) \(7^3 = 343\) becomes \(\log_7(343) = 3\). 3. b) \(2^{-4} = \frac{1}{16}\) becomes \(\log_2\left(\frac{1}{16}\right) = -4\). 4. c) \(\log_3(243) = 5\) becomes \(3^5 = 243\). 5. d) Since \(0.001 = 10^{-3}\), the equation \(10^x = 0.001\) gives \(x = -3\). Its logarithmic form is \(\log_{10}(0.001) = x\). 6. e) The equation becomes \(2^{-3} = x\), so \(x = \frac{1}{8} = 0.125\).

a) \(\log_7(343) = 3\) b) \(\log_2\left(\frac{1}{16}\right) = -4\) c) \(3^5 = 243\) d) \(\log_{10}(0.001) = x\) and \(x = -3\) e) \(2^{-3} = x\) and \(x = \frac{1}{8}\)

- Convert each logarithmic statement to exponential form. - Interpret a negative exponent as a reciprocal. - Express \(16\) and \(4\) as powers of the same base. - Rewrite the square root using a rational exponent.

1. a) \(x = 2^{-4} = \frac{1}{16}\). 2. b) \(b^{-2} = 0.01 = \frac{1}{100}\), so \(b^2 = 100\). A logarithm base must be positive, so \(b = 10\). 3. c) Let \(\log_{16}(4) = y\). Then \(16^y = 4\). Since \(16 = 4^2\), \(4^{2y} = 4^1\), so \(y = \frac{1}{2}\). 4. d) Since \(125 = 5^3\), \(\sqrt{125} = 5^{3/2}\). Therefore, \(y = \frac{3}{2}\).

a) \(x = \frac{1}{16}\) b) \(b = 10\) c) \(\frac{1}{2}\) d) \(y = \frac{3}{2}\)

- Rewrite each argument as a power of \(2\). - What does a negative exponent mean? - Express roots using fractional exponents. - Identify the exponent that makes \(2^x\) equal to the given argument.

1. Since \(2^5 = 32\), \(\log_2(32) = 5\). 2. Rewrite \(\frac{1}{8}\) as \(2^{-3}\). Therefore, \(\log_2\left(\frac{1}{8}\right) = -3\). 3. Rewrite \(\sqrt[3]{2}\) as \(2^{1/3}\). Therefore, \(\log_2(\sqrt[3]{2}) = \frac{1}{3}\). 4. Rewrite \(\frac{1}{\sqrt{2}}\) as \(2^{-1/2}\). Therefore, \(\log_2\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2}\).

1) \(5\) 2) \(-3\) 3) \(\frac{1}{3}\) 4) \(-\frac{1}{2}\)

- Rewrite each logarithmic equation as an exponential equation. - A negative exponent creates a reciprocal. - A unit-fraction exponent represents a root. - What is the value of a nonzero base raised to the zeroth power?

1. Use \(\log_a(b) = c\) if and only if \(b = a^c\). 2. a) \(b = (0.5)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\). 3. b) \(b = 2^{-5} = \frac{1}{32}\). 4. c) \(b = 8^{1/3} = \sqrt[3]{8} = 2\). 5. d) \(b = 11^0 = 1\).

a) \(b = \frac{1}{8}\) b) \(b = \frac{1}{32}\) c) \(b = 2\) d) \(b = 1\)

- Rewrite the argument as a power of the logarithm base. - Convert a terminating decimal to a fraction when helpful. - A common logarithm has base \(10\). - Simplify radicals before evaluating the logarithm.

1. a) Since \(7^2 = 49\), \(\log_7(49) = 2\). 2. b) Since \(\frac{1}{7} = 7^{-1}\), \(\log_7\left(\frac{1}{7}\right) = -1\). 3. c) Rewrite \(0.125\) as \(\frac{1}{8} = 2^{-3}\). Therefore, \(\log_2(0.125) = -3\). 4. d) Since \(0.01 = 10^{-2}\), \(\log(0.01) = -2\). 5. e) First, \(\sqrt{64} = 8\). Since \(4^{3/2} = 8\), \(\log_4(\sqrt{64}) = \frac{3}{2}\).

a) \(2\) b) \(-1\) c) \(-3\) d) \(-2\) e) \(\frac{3}{2}\)

- Ask which exponent on the base produces the argument. - Rewrite reciprocals using negative exponents. - Recall the value of a logarithm whose argument is \(1\). - List a few powers of \(2\), \(3\), \(5\), and \(7\) if needed.

1. a) Since \(2^7 = 128\), \(\log_2(128) = 7\). 2. b) Since \(\frac{1}{25} = 5^{-2}\), \(\log_5\left(\frac{1}{25}\right) = -2\). 3. c) Since \(3^5 = 243\), \(\log_3(243) = 5\). 4. d) Since \(9^0 = 1\), \(\log_9(1) = 0\). 5. e) Since \(7^3 = 343\), \(\log_7(343) = 3\).

a) \(7\) b) \(-2\) c) \(5\) d) \(0\) e) \(3\)

- How are logarithmic and exponential forms related? - What does a negative exponent mean? - What is the value of any nonzero base raised to the zeroth power? - Rewrite a rational exponent as a root.

1. a) Rewrite in exponential form: \(x = 10^4 = 10{,}000\). 2. b) Rewrite in exponential form: \(x = 2^{-3} = \frac{1}{2^3} = \frac{1}{8} = 0.125\). 3. c) Rewrite in exponential form: \(x = 7^0 = 1\). 4. d) Rewrite in exponential form: \(x = 25^{1/2} = \sqrt{25} = 5\).

a) \(x = 10{,}000\) b) \(x = \frac{1}{8}\) c) \(x = 1\) d) \(x = 5\)

- Rewrite each logarithmic equation in exponential form. - What base is understood when \(\log\) denotes the common logarithm? - Write a decimal exponent as a fraction before evaluating the power. - Recall how a negative exponent changes a power.

1. a) Rewrite in exponential form: \(x = 3^4 = 81\). 2. b) Rewrite in exponential form: \(x = 5^{-2} = \frac{1}{5^2} = \frac{1}{25} = 0.04\). 3. c) Write the decimal exponent as a fraction: \(x = 9^{2.5} = 9^{5/2} = (\sqrt{9})^5 = 3^5 = 243\). 4. d) A common logarithm has base \(10\), so \(x = 10^{-1} = \frac{1}{10} = 0.1\).

a) \(x = 81\) b) \(x = 0.04\) c) \(x = 243\) d) \(x = 0.1\)

- Ask what exponent on the base produces the logarithm's argument. - Rewrite roots and reciprocals using rational and negative exponents. - Apply the identity \(\log_b(b^r) = r\).

1. Use the inverse identity \(\log_b(b^r) = r\). 2. a) The argument is already \(7^n\), so \(\log_7(7^n) = n\). 3. b) Rewrite \(\frac{1}{3^k}\) as \(3^{-k}\). Then \(\log_3(3^{-k}) = -k\). 4. c) Rewrite \(\sqrt[m]{2}\) as \(2^{1/m}\). Then \(\log_2(2^{1/m}) = \frac{1}{m}\). 5. d) Rewrite \(\frac{1}{\sqrt[4]{a^3}}\) as \(a^{-3/4}\). Then \(\log_a(a^{-3/4}) = -\frac{3}{4}\).

a) \(n\) b) \(-k\) c) \(\frac{1}{m}\) d) \(-\frac{3}{4}\)

- Use the definition of a logarithm as the inverse of exponentiation. - Rewrite \(\log_a(u) = v\) as \(a^v = u\). - Recall that an exponent of \(0.5\) represents a square root. - Rewrite a negative exponent using a reciprocal.

1. Use \(\log_a(u) = v\) if and only if \(a^v = u\). 2. a) \(a^3 = 216\), so \(a = \sqrt[3]{216} = 6\). 3. b) \(a^{1/2} = 5\), so \(\sqrt{a} = 5\) and \(a = 25\). 4. c) \(a^{-5} = \frac{1}{32} = 2^{-5}\). Since valid logarithm bases are positive, \(a = 2\). 5. d) \(a^{1/2} = \sqrt{11}\), so \(\sqrt{a} = \sqrt{11}\) and \(a = 11\).

a) \(a = 6\) b) \(a = 25\) c) \(a = 2\) d) \(a = 11\)

- Interpret \(\log_b(u)\) as the exponent on \(b\) that produces \(u\). - Exponentiation with base \(b\) reverses a logarithm with the same base. - Look for the pattern \(b^{\log_b(u)}\).

Use the inverse identity \(b^{\log_b(u)} = u\). 1. a) \(12^{\log_{12}(8)} = 8\). 2. b) Because the common logarithm has base \(10\), \(10^{\log(5a)} = 5a\). 3. c) \(2^{\log_2(\sqrt{11})} = \sqrt{11}\). 4. d) \(x^{\log_x(25)} = 25\).

a) \(8\) b) \(5a\) c) \(\sqrt{11}\) d) \(25\)

- Interpret each logarithm in terms of powers of its base. - Find one power just below and one power just above the argument. - For an argument between \(0\) and \(1\), consider negative exponents. - For the bases used here, larger exponents produce larger powers.

1. a) Since \(2^4 = 16\) and \(2^5 = 32\), \(16 < 25 < 32\). Therefore, \(4 < \log_2(25) < 5\). 2. b) Since \(3^{-3} = \frac{1}{27}\) and \(3^{-2} = \frac{1}{9}\), \(\frac{1}{27} < \frac{1}{10} < \frac{1}{9}\). Therefore, \(-3 < \log_3\left(\frac{1}{10}\right) < -2\). 3. c) Since \(10^{-2} = 0.01\) and \(10^{-1} = 0.1\), \(0.01 < 0.05 < 0.1\). Therefore, \(-2 < \log(0.05) < -1\).

a) \(4 < \log_2(25) < 5\) b) \(-3 < \log_3\left(\frac{1}{10}\right) < -2\) c) \(-2 < \log(0.05) < -1\)

- Apply the inverse exponential function with base \(9\). - Interpret an exponent of \(0.5\) as a square root. - Decide whether a base greater than \(1\) preserves or reverses inequalities. - Check the logarithm domain.

1. Apply the exponential function with base \(9\) to all parts of the compound inequality: \(9^{0.5} \le 9^{\log_9(x)} < 9^1\). 2. Since the base \(9\) is greater than \(1\), the inequality directions remain unchanged. 3. Simplify: \(3 \le x < 9\). 4. Every value in this interval is positive, so the logarithm domain condition is satisfied.

\([3, 9)\)

- Require the logarithm argument to be positive. - Apply the inverse exponential function with base \(2\). - Determine whether the inequality direction changes. - Intersect the result with the domain.

1. The logarithm requires \(x+4 > 0\), so \(x > -4\). 2. Because the base \(2\) is greater than \(1\), exponentiating preserves the inequality: \(x+4 \le 2^3\). 3. Thus, \(x+4 \le 8\), so \(x \le 4\). 4. Intersect this result with the domain: \(-4 < x \le 4\).

\((-4, 4]\)

- First identify the base, exponent, and power in the given equation. - The logarithm's base is the same as the exponential expression's base. - A logarithm asks which exponent on the base produces the given value. - The \(b\)th root of the power returns the positive base \(a\).

Identify the base \(a\), exponent \(b\), and power \(c\) in the equivalence \(a^b = c\), \(\sqrt[b]{c} = a\), and \(\log_a(c) = b\). 1. a) Here \(a = 7\), \(b = 3\), and \(c = 343\). The other forms are \(\sqrt[3]{343} = 7\) and \(\log_7(343) = 3\). 2. b) Here \(a = 2\), \(b = 6\), and \(c = 64\). The other forms are \(2^6 = 64\) and \(\sqrt[6]{64} = 2\). 3. c) Here \(a = 5\), \(b = 4\), and \(c = 625\). The other forms are \(5^4 = 625\) and \(\log_5(625) = 4\). 4. d) Here \(a = \frac{1}{2}\), \(b = 4\), and \(c = 0.0625\). The other forms are \(\sqrt[4]{0.0625} = \frac{1}{2}\) and \(\log_{1/2}(0.0625) = 4\).

a) \(\sqrt[3]{343} = 7\); \(\log_7(343) = 3\) b) \(2^6 = 64\); \(\sqrt[6]{64} = 2\) c) \(5^4 = 625\); \(\log_5(625) = 4\) d) \(\sqrt[4]{0.0625} = \frac{1}{2}\); \(\log_{1/2}(0.0625) = 4\)

- A common logarithm has base \(10\). - Convert each logarithmic equation to exponential form. - Rewrite a decimal exponent as a fraction. - Interpret a negative exponent as a reciprocal.

1. a) Since a common logarithm has base \(10\), \(x = 10^5 = 100{,}000\). 2. b) \(x = (0.1)^{-3} = \left(\frac{1}{10}\right)^{-3} = 10^3 = 1000\). 3. c) \(x = 4^{1.5} = 4^{3/2} = (\sqrt{4})^3 = 8\).

a) \(x = 100{,}000\) b) \(x = 1000\) c) \(x = 8\)

- Convert each logarithmic equation to exponential form. - Identify the positive number that produces the argument when raised to the given exponent. - A negative exponent creates a reciprocal. - Apply the restrictions on a logarithm base.

1. Rewrite \(\log_x(u) = v\) as \(x^v = u\). 2. a) \(x^4 = 625\). A logarithm base must be positive, so \(x = 5\). 3. b) \(x^{-5} = \frac{1}{32}\), so \(x^5 = 32\) and \(x = 2\). 4. c) \(x^{-4} = 0.0001 = \frac{1}{10000}\), so \(x^4 = 10{,}000\). The positive base is \(x = 10\). 5. d) \(x^1 = 6\), so \(x = 6\).

a) \(x = 5\) b) \(x = 2\) c) \(x = 10\) d) \(x = 6\)

- Relate logarithms to exponential equations and roots. - Isolate the base in an equation of the form \(b^n = a\). - Interpret a fractional exponent as a root. - Convert a terminating decimal to a fraction when helpful.

1. Use \(\log_b(u) = v\) if and only if \(b^v = u\). 2. a) \(b^3 = 343\), so \(b = 7\). 3. b) \(b^{-2} = \frac{1}{9}\), so \(b^2 = 9\). A logarithm base must be positive, so \(b = 3\). 4. c) \(b^{1/3} = 2\), so \(b = 2^3 = 8\). 5. d) \(b^{-1} = 0.125 = \frac{1}{8}\), so \(b = 8\).

a) \(b = 7\) b) \(b = 3\) c) \(b = 8\) d) \(b = 8\)

- Ask which exponent on the base produces the argument. - Rewrite the argument as a power of the stated base. - Express reciprocals and roots using negative or fractional exponents. - A common logarithm has base \(10\).

1. a) Since \(2^9 = 512\), \(\log_2(512) = 9\). 2. b) Since \(\frac{1}{27} = 3^{-3}\), \(\log_3\left(\frac{1}{27}\right) = -3\). 3. c) Since \(0.01 = 10^{-2}\), \(\log(0.01) = -2\). 4. d) Since \(4 = 16^{1/2}\), \(\log_{16}(4) = \frac{1}{2}\). 5. e) Since \(\sqrt[3]{5} = 5^{1/3}\), \(\log_5(\sqrt[3]{5}) = \frac{1}{3}\).

a) \(9\) b) \(-3\) c) \(-2\) d) \(\frac{1}{2}\) e) \(\frac{1}{3}\)

- How can you rewrite each root as a fractional exponent? - Express the logarithm's argument as a power of its base. - Think of a reciprocal as a negative power. - What exponent gives the argument when applied to the given base?

1. Rewrite \(\sqrt[3]{16}=(2^4)^{1/3}=2^{4/3}\), so the logarithm is \(\frac{4}{3}\). 2. Rewrite \(\sqrt[4]{125}=(5^3)^{1/4}=5^{3/4}\), so the logarithm is \(\frac{3}{4}\). 3. Rewrite \(\frac{1}{\sqrt{27}}=(3^3)^{-1/2}=3^{-3/2}\), so the logarithm is \(-\frac{3}{2}\). 4. Rewrite \(\sqrt[5]{0.1}=(10^{-1})^{1/5}=10^{-1/5}\), so the logarithm is \(-\frac{1}{5}\).

a) \(\frac{4}{3}\) b) \(\frac{3}{4}\) c) \(-\frac{3}{2}\) d) \(-\frac{1}{5}\)

- Find two consecutive integer powers of the base that bracket the argument. - Negative exponents produce positive values between \(0\) and \(1\) when the base is greater than \(1\). - Use the relationship between logarithmic and exponential form.

1. a) Since \(3^3 = 27\) and \(3^4 = 81\), \(27 < 80 < 81\). Therefore, \(3 < \log_3(80) < 4\). 2. b) Since \(2^{-1} = 0.5\) and \(2^0 = 1\), \(0.5 < 0.7 < 1\). Therefore, \(-1 < \log_2(0.7) < 0\). 3. c) Since \(4^3 = 64\) and \(4^4 = 256\), \(64 < 70 < 256\). Therefore, \(3 < \log_4(70) < 4\). 4. d) Since \(10^2 = 100\) and \(10^3 = 1000\), \(100 < 500 < 1000\). Therefore, \(2 < \log_{10}(500) < 3\).

a) Between \(3\) and \(4\) b) Between \(-1\) and \(0\) c) Between \(3\) and \(4\) d) Between \(2\) and \(3\)

- Find two consecutive powers of each base with the argument between them. - Which negative exponents give numbers between \(0\) and \(1\)? - Convert the inequalities about powers into inequalities about exponents.

1. Since \(5^{-3}=0.008<0.03<0.04=5^{-2}\), the bounds are \(-3\) and \(-2\). 2. Since \(8^2=64<100<512=8^3\), the bounds are \(2\) and \(3\). 3. Since \(2^1=2<2.5<4=2^2\), the bounds are \(1\) and \(2\). 4. Since \(10^{-3}=0.001<0.002<0.01=10^{-2}\), the bounds are \(-3\) and \(-2\).

a) \(-3<\log_5(0.03)<-2\) b) \(2<\log_8(100)<3\) c) \(1<\log_2(2.5)<2\) d) \(-3<\log_{10}(0.002)<-2\)

- Simplify each expression with a logarithm in the exponent before solving the equation. - What base is used for a common logarithm? - Replace the exponential-logarithmic expressions with their simplified values, then solve the resulting linear equation.

1. Use \(b^{\log_b(u)} = u\) to simplify each exponential expression: \(5^{\log_5(4)} = 4\), \(10^{\log(31)} = 31\), and \(2^{\log_2(9)} = 9\). 2. Substitute these values into the equation: \(4a = 31 + 9\). 3. The right side equals \(40\), so \(4a = 40\). 4. Divide by \(4\): \(a = 10\).

\(a = 10\)

- Substitute the point’s coordinates into the function equation. - Rewrite each logarithmic equation in exponential form. - Use a root to solve an equation such as \(b^3 = 216\). - Recall how a negative exponent creates a reciprocal.

1. If \((x, y)\) lies on \(y = \log_b(x)\), then \(b^y = x\). 2. a) \(b^3 = 216\), so \(b = \sqrt[3]{216} = 6\). 3. b) \(b^{-3} = 0.125 = \frac{1}{8}\). Thus, \(b^3 = 8\), so \(b = 2\). 4. c) \(b^1 = 5\), so \(b = 5\).

a) \(b = 6\) b) \(b = 2\) c) \(b = 5\)

- Substitute each point’s coordinates into the function equation. - Rewrite the logarithmic equation in exponential form. - Interpret an exponent of \(0.5\) as a square root. - Convert terminating decimals to fractions when helpful.

1. Use \(y = \log_b(x)\) if and only if \(b^y = x\). 2. a) \(b^{0.5} = 10\), so \(\sqrt{b} = 10\). Squaring gives \(b = 100\). 3. b) \(b^{-4} = \frac{1}{81}\), so \(b^4 = 81\). Because a logarithm base must be positive, \(b = 3\). 4. c) \(b^{-2} = 0.04 = \frac{1}{25}\), so \(b^2 = 25\). Because the base must be positive, \(b = 5\).

a) \(b = 100\) b) \(b = 3\) c) \(b = 5\)

- How are logarithmic and exponential equations related? - Substitute the coordinates of \(P\) into the function and rewrite the equation in exponential form. - Which positive number raised to the fifth power equals \(32\)? - Use the base you found to evaluate the second logarithm.

1. Substitute the point into the function: \(5 = \log_a(32)\). 2. Rewrite in exponential form: \(a^5 = 32\). 3. Since \(2^5 = 32\), the base is \(a = 2\). 4. Then \(f(8) = \log_2(8)\). Because \(2^3 = 8\), \(f(8) = 3\).

The base is \(a = 2\), and \(f(8) = 3\).

- Use \(y=\log_b(x)\) if and only if \(b^y=x\). - Substitute the point's coordinates into the exponential form. - Solve for \(b\), remembering that a logarithm base must be positive and cannot equal \(1\).

1. A point \((x, y)\) on \(y=\log_b(x)\) satisfies \(b^y=x\). 2. For a), \(b^2=100\). Since \(b>0\), \(b=10\). 3. For b), \(b^{-2}=\frac{1}{9}\), so \(b^2=9\). Since \(b>0\), \(b=3\). 4. For c), \(b^{0.5}=2\), so \(\sqrt{b}=2\) and \(b=4\). 5. For d), \(b^1=5\), so \(b=5\).

a) \(b=10\) b) \(b=3\) c) \(b=4\) d) \(b=5\)

- What happens to the input and output coordinates when a function is inverted? - Write a pair for each table row, then reverse the coordinates. - Which function undoes raising \(3\) to a power?

1. Swapping the coordinates of every \((x, f(x))\) pair gives points on the inverse. 2. The inverse points are \((\frac{1}{9}, -2)\), \((\frac{1}{3}, -1)\), \((1, 0)\), \((3, 1)\), and \((9, 2)\). 3. The operation that undoes \(3^x\) is \(\log_3(x)\), so \(f^{-1}(x)=\log_3(x)\).

Points: \((\frac{1}{9}, -2)\), \((\frac{1}{3}, -1)\), \((1, 0)\), \((3, 1)\), \((9, 2)\). Rule: \(f^{-1}(x)=\log_3(x)\).

- Compare the coordinates of the labeled points. - What geometric change takes one input-output pair to its inverse pair? - What operation reverses raising a number to a power?

1. Point \(A=(1, 2)\) on \(f\) corresponds to \(B=(2, 1)\) on \(g\); their coordinates are reversed. 2. The curves reflect across \(y=x\), the mirror line for inverses. 3. The inverse of \(f(x)=2^x\) is \(g(x)=\log_2(x)\).

\(g(x)=\log_2(x)\). The graphs reflect across \(y=x\); \(A=(1, 2)\) corresponds to \(B=(2, 1)\).

- Find the two requested x-values on their respective curves, then read the heights. - How do the two input-output pairs compare? - Which diagonal line swaps the roles of the axes under reflection?

1. The exponential graph gives \(f(2)=4\). The inverse graph gives \(g(4)=2\). 2. The points \((2,4)\) and \((4,2)\) exchange coordinates, as inverse functions do. Their graphs reflect across \(y=x\).

a) \(f(2)=4\) and \(g(4)=2\) b) The input-output pair is reversed; the reflection line is \(y=x\).

- Write the square root as a rational exponent. - Ask what exponent on the base produces the logarithm's argument. - Can a positive base raised to a real power produce a negative result? - Rewrite the logarithmic equation in exponential form.

1. \(\log_2(64)=6\) because \(2^6=64\). Also, \(\sqrt{10}=10^{1/2}\), so \(\log_{10}(\sqrt{10})=\frac{1}{2}\). 2. \(\log_b(b^5)=5\). The logarithm asks for the exponent on \(b\) that produces \(b^5\), and that exponent is \(5\). 3. The equation \(\log_2(-8)=x\) is equivalent to \(2^x=-8\). Since \(2^x>0\) for every real \(x\), the equation has no real solution. Therefore, a real logarithm cannot have a negative argument.

a) \(\log_2(64)=6\) and \(\log_{10}(\sqrt{10})=\frac{1}{2}\) b) \(\log_b(b^5)=5\) c) No real solution; \(2^x\) is always positive for real \(x\).

- Rewrite each argument as a power of its logarithm base. - Express the square root using a rational exponent. - What logarithm value corresponds to an argument of \(1\)? - Compare the resulting exact numbers on a number line.

1. Since \(\frac{1}{9} = 3^{-2}\), \(P = \log_3(3^{-2}) = -2\). 2. Since \(\sqrt{8} = (2^3)^{1/2} = 2^{3/2}\), \(Q = \log_2(2^{3/2}) = \frac{3}{2}\). 3. Since \(5^0 = 1\), \(R = \log_5(1) = 0\). 4. Compare the values: \(-2 < 0 < \frac{3}{2}\). Therefore, \(P < R < Q\).

\(P = -2\), \(Q = \frac{3}{2}\), and \(R = 0\) Least to greatest: \(P < R < Q\)

- What values are allowed as the argument of a logarithm? - Use the definition of a logarithm to find the exponent that produces each input. - What transformation occurs when every function value is multiplied by \(-1\)? - Think about the behavior of a logarithmic function with a base greater than \(1\) and locate its x-intercept.

1. The argument of a logarithm must be positive, so both functions have domain \(\{x \in \mathbb{R} \mid x > 0\}\). 2. For \(f(x) = \log_2(x)\): \(f(0.25) = -2\), \(f(1) = 0\), \(f(2) = 1\), and \(f(4) = 2\). For \(g(x) = \log_{0.5}(x)\): \(g(0.25) = 2\), \(g(1) = 0\), \(g(2) = -1\), and \(g(4) = -2\). 3. Since \(0.5 = \frac{1}{2}\), \(g(x) = \log_{1/2}(x) = -\log_2(x) = -f(x)\). Multiplying all output values by \(-1\) reflects a graph across the x-axis. 4. The function \(f\) is increasing because its base is greater than \(1\). Since \(f(1) = 0\) and \(0.75 < 1\), it follows that \(f(0.75) < 0\).

a) Both domains are \(\{x \in \mathbb{R} \mid x > 0\}\). b) \(f(0.25) = -2\), \(f(1) = 0\), \(f(2) = 1\), \(f(4) = 2\); \(g(0.25) = 2\), \(g(1) = 0\), \(g(2) = -1\), \(g(4) = -2\) c) \(g(x) = -f(x)\), so the graph of \(g\) is the reflection of the graph of \(f\) across the x-axis. d) Negative, because \(f\) is increasing and \(0.75 < 1\), where \(f(1) = 0\).

- Identify consecutive powers of the base that bound the argument. - A logarithmic function with a base greater than \(1\) is increasing. - For an argument between \(0\) and \(1\), consider negative exponents. - A common logarithm has base \(10\).

1. a) Since \(3^2 = 9\) and \(3^3 = 27\), \(9 < 20 < 27\). The base-\(3\) logarithm is increasing, so \(2 < \log_3(20) < 3\). 2. b) Since \(2^{-2} = 0.25\) and \(2^{-1} = 0.5\), \(0.25 < 0.3 < 0.5\). The base-\(2\) logarithm is increasing, so \(-2 < \log_2(0.3) < -1\). 3. c) Since \(10^2 = 100\) and \(10^3 = 1000\), \(100 < 500 < 1000\). Therefore, \(2 < \log(500) < 3\).

a) \(2 < \log_3(20) < 3\) b) \(-2 < \log_2(0.3) < -1\) c) \(2 < \log(500) < 3\)

- State the domain restriction for each logarithm's argument. - Because the common logarithm is increasing, exponentiating with base \(10\) preserves the inequality direction. - A common logarithm has base \(10\). - Intersect the inequality solution with the logarithm's domain.

1. The domain requires \(x > 0\). Since the common logarithm is increasing, \(\log(x) < 2\) is equivalent to \(x < 10^2 = 100\). Thus, \(0 < x < 100\). 2. The domain requires \(2x > 0\), so \(x > 0\). Since the logarithm is increasing, \(\log(2x) \ge 1\) gives \(2x \ge 10\), so \(x \ge 5\). This already satisfies the domain restriction. 3. The domain requires \(x - 1 > 0\), so \(x > 1\). The inequality \(\log(x - 1) < 0\) gives \(x - 1 < 10^0 = 1\), so \(x < 2\). Combining the restrictions gives \(1 < x < 2\).

1) \(0 < x < 100\) 2) \(x \ge 5\) 3) \(1 < x < 2\)

- How does an exponential function behave when its base is greater than \(1\)? - How does it behave when its base is between \(0\) and \(1\)? - Compare \(b^x\) with \(b^1\) when \(x > 1\). - Can an exponential function with a positive base have an output of \(0\) or a negative output?

1. a) For \(b = 2\), the exponential function \(2^x\) is increasing. If \(x > 1\), then \(y = 2^x > 2^1 = 2\). Therefore, \(y \in (2, \infty)\). 2. b) For \(b = 0.5\), the exponential function \((0.5)^x\) is decreasing. If \(x > 1\), then \(y = (0.5)^x < (0.5)^1 = 0.5\). Exponential outputs are positive, so \(y \in (0, 0.5)\). 3. When the base is greater than \(1\), larger exponents produce larger outputs. When the base is between \(0\) and \(1\), larger exponents produce smaller positive outputs.

a) \(y \in (2, \infty)\) b) \(y \in (0, 0.5)\) The intervals differ because the base-\(2\) function is increasing while the base-\(0.5\) function is decreasing.

- Express each radical using a fractional exponent. - Think about what exponent the logarithm is asking you to find. - After simplifying the two logarithms, how do you divide their values?

1. Rewrite the numerator argument: \(a^2\sqrt{a}=a^{5/2}\). Therefore, \(\log_a(a^{5/2})=\frac{5}{2}\). 2. Rewrite the denominator argument: \(\sqrt[3]{a^2}=a^{2/3}\). Therefore, \(\log_a(a^{2/3})=\frac{2}{3}\). 3. Divide the two values: \(x=\frac{5/2}{2/3}=\frac{15}{4}\).

\(x=\frac{15}{4}\)

- A common logarithm has base \(10\). - Rewrite each quantity as a power of a convenient common base. - Express \(0.5\) as \(\frac{1}{2}\) or \(2^{-1}\). - Rewrite a root in a denominator using a negative rational exponent.

1. a) Since \(0.001 = 10^{-3}\), \(\log(0.001) = -3\). 2. b) Since \(1000 = 10^3\), \(\sqrt[4]{1000} = 10^{3/4}\). Therefore, \(\log(\sqrt[4]{1000}) = \frac{3}{4}\). 3. c) Let \(\log_{0.5}(8) = x\). Since \(0.5 = 2^{-1}\) and \(8 = 2^3\), \(2^{-x} = 2^3\), so \(x = -3\). 4. d) Rewrite the argument as \(2^{-1/3}\). If the logarithm equals \(x\), then \((2^{-1})^x = 2^{-1/3}\). Thus, \(-x = -\frac{1}{3}\), so \(x = \frac{1}{3}\).

a) \(-3\) b) \(\frac{3}{4}\) c) \(-3\) d) \(\frac{1}{3}\)

- Rewrite the quantity under each radical as a power of the logarithm base. - Express an \(n\)th root using an exponent of \(\frac{1}{n}\). - A factor in the denominator contributes a negative exponent. - Add exponents when multiplying powers with the same base.

1. a) \(\sqrt[4]{27} = \sqrt[4]{3^3} = 3^{3/4}\), so the logarithm equals \(\frac{3}{4}\). 2. b) \(\frac{1}{\sqrt[5]{8}} = \frac{1}{(2^3)^{1/5}} = 2^{-3/5}\), so the logarithm equals \(-\frac{3}{5}\). 3. c) \(\sqrt[3]{100} = \sqrt[3]{10^2} = 10^{2/3}\), so the common logarithm equals \(\frac{2}{3}\). 4. d) \(25\sqrt{5} = 5^2 \cdot 5^{1/2} = 5^{5/2}\), so the logarithm equals \(\frac{5}{2}\). 5. e) \(\frac{1}{49\sqrt[3]{7}} = 7^{-(2+1/3)} = 7^{-7/3}\), so the logarithm equals \(-\frac{7}{3}\).

a) \(\frac{3}{4}\) b) \(-\frac{3}{5}\) c) \(\frac{2}{3}\) d) \(\frac{5}{2}\) e) \(-\frac{7}{3}\)

- Convert terminating decimals to fractions first. - When the base does not match directly, relate \(2\) and \(4\) as powers of the same number. - Divide powers with the same base by subtracting exponents. - Express each root using a rational exponent.

1. a) Since \(0.125 = \frac{1}{8} = 2^{-3}\), \(\sqrt{0.125} = 2^{-3/2}\). Therefore, the value is \(-\frac{3}{2}\). 2. b) Since \(0.04 = \frac{1}{25} = 5^{-2}\), \(\sqrt[4]{0.04} = 5^{-1/2}\). Therefore, the value is \(-\frac{1}{2}\). 3. c) Rewrite the argument as \(\frac{10^1}{(10^2)^{1/5}} = 10^{1-2/5} = 10^{3/5}\). Therefore, the common logarithm is \(\frac{3}{5}\). 4. d) Rewrite the argument as \(\frac{3^{3/2}}{3^2} = 3^{-1/2}\). Therefore, the value is \(-\frac{1}{2}\). 5. e) Since \(32 = 2^5\), \(\sqrt[3]{32} = 2^{5/3} = 4^{5/6}\). Therefore, the value is \(\frac{5}{6}\).

a) \(-\frac{3}{2}\) b) \(-\frac{1}{2}\) c) \(\frac{3}{5}\) d) \(-\frac{1}{2}\) e) \(\frac{5}{6}\)

- A point's coordinates give an input and its corresponding function value. - Substitute the point into the function equation. - Rewrite \(y = \log_a(x)\) as \(a^y = x\). - Express negative exponents and decimals as reciprocals or fractions when useful.

1. Substitute each point \((x, y)\) into \(y = \log_a(x)\), then rewrite the equation as \(a^y = x\). 2. a) \(a^4 = 16\), so \(a = 2\). 3. b) \(a^{-2} = 0.01 = \frac{1}{100}\), so \(a^2 = 100\). A logarithm base is positive, so \(a = 10\). 4. c) \(a^{1/2} = 3\), so \(\sqrt{a} = 3\) and \(a = 9\). 5. d) \(a^{-1} = 2\), so \(\frac{1}{a} = 2\) and \(a = \frac{1}{2}\).

a) \(a = 2\) b) \(a = 10\) c) \(a = 9\) d) \(a = \frac{1}{2}\)

- First use exponent rules to rewrite each argument as one power of the logarithm's base. - Rewrite each root using a rational exponent. - What happens to exponents when powers with the same base are multiplied or divided?

1. a) Combine powers with the same base: \(x^2 \cdot x^5 = x^7\). Therefore, \(\log_x(x^7) = 7\). 2. b) Rewrite the square root as a rational exponent: \(\frac{\sqrt{b}}{b^3} = b^{1/2-3} = b^{-5/2}\). Therefore, the logarithm equals \(-\frac{5}{2}\). 3. c) Rewrite the cube root as \(\sqrt[3]{c^2} = c^{2/3}\). Therefore, the logarithm equals \(\frac{2}{3}\). 4. d) Rewrite the fifth root as \(\sqrt[5]{d} = d^{1/5}\). Then \(\frac{d}{\sqrt[5]{d}} = d^{1-1/5} = d^{4/5}\), so the logarithm equals \(\frac{4}{5}\).

a) \(7\) b) \(-\frac{5}{2}\) c) \(\frac{2}{3}\) d) \(\frac{4}{5}\)

- Rewrite each logarithmic equation in exponential form. - Recall how a negative exponent creates a reciprocal. - Express terminating decimals as powers or fractions when helpful. - Write \(1.5\) as \(\frac{3}{2}\) before solving for the base.

1. Convert each equation to exponential form. 2. a) \(a^{-3} = 0.001 = 10^{-3}\), so \(a = 10\). 3. b) \(a^{3/2} = 8\). Raise both sides to the power \(\frac{2}{3}\): \(a = 8^{2/3} = (\sqrt[3]{8})^2 = 4\). 4. c) \(a^{-1} = \frac{1}{7} = 7^{-1}\), so \(a = 7\). 5. d) \(a^{1/3} = 2\). Cubing both sides gives \(a = 2^3 = 8\).

a) \(a = 10\) b) \(a = 4\) c) \(a = 7\) d) \(a = 8\)

- Simplify the exponential expression containing a logarithm before performing the remaining operations. - Use the identity \(a^{\log_a(u)} = u\). - After simplifying the logarithmic part, follow the usual order of operations.

1. a) First, \(3^{\log_3(4)} = 4\). Then \(5 \cdot 4 - 7 = 20 - 7 = 13\). 2. b) Since the common logarithm has base \(10\), \(10^{\log(8)} = 8\) and \(10^{\log(12)} = 12\). Their sum is \(20\). 3. c) The numerator simplifies to \(6^{\log_6(18)} = 18\). Then \(18 \div 3 = 6\). 4. d) The expression inside the parentheses is \(4^{\log_4(5)} = 5\). Therefore, the value is \(5^2 = 25\).

a) \(13\) b) \(20\) c) \(6\) d) \(25\)

- Estimate each logarithm separately using nearby powers. - Find consecutive powers of each base that surround the argument. - Decide whether the integer bounds alone are enough to compare the three values.

1. Since \(2^5 = 32\) and \(2^6 = 64\), \(32 < 40 < 64\), so \(5 < x < 6\). 2. Since \(3^3 = 27\) and \(3^4 = 81\), \(27 < 80 < 81\), so \(3 < y < 4\). 3. Since \(5^2 = 25\) and \(5^3 = 125\), \(25 < 120 < 125\), so \(2 < z < 3\). 4. Because \(x > 5\), while \(y < 4\) and \(z < 3\), \(x\) is the greatest value.

\(x = \log_2(40)\) is greatest because \(5 < x < 6\), \(3 < y < 4\), and \(2 < z < 3\).

- Rewrite each logarithmic equation in exponential form. - When the base is unknown, solve the resulting power equation for the base. - Rewrite roots and reciprocals using fractional or negative exponents. - Use the inverse relationship between logarithms and exponential functions.

1. a) Rewrite the equation in exponential form: \(x^3 = 1000\). Therefore, \(x = 10\). 2. b) Rewrite the equation in exponential form: \(x = 4^{2.5} = 4^{5/2} = 32\). 3. c) The equation is \(\left(\frac{1}{3}\right)^x = 9\). Since \(\left(\frac{1}{3}\right)^{-2} = 9\), \(x = -2\). 4. d) Rewrite the argument as \(\sqrt{7^5} = 7^{5/2}\). Therefore, \(x = \frac{5}{2}\). 5. e) Use the inverse relationship \(\log_b(b^y) = y\). Thus, \(x = k-1\).

a) \(x = 10\) b) \(x = 32\) c) \(x = -2\) d) \(x = \frac{5}{2}\) e) \(x = k-1\)

- What happens when a value is first evaluated with a logarithm and then used as an exponent with the same base? - Look for expressions of the form \(b^{\log_b(u)}\). - Use the relationship between squaring and taking the principal square root. - Simplify the two terms separately before combining them.

1. Apply \(b^{\log_b(u)} = u\) to the expression inside the parentheses: \(10^{\log(\sqrt{x})} = \sqrt{x}\). 2. Since \(x > 0\), \((\sqrt{x})^2 = x\). 3. Apply the same inverse identity to the second term: \(8^{\log_8(3x)} = 3x\). 4. Combine like terms: \(x + 3x = 4x\).

\(4x\)

- What happens to the roles of inputs and outputs when a function is inverted? - How are the domain of a function and the range of its inverse related? - If \((a, b)\) lies on a function's graph, which ordered pair lies on the inverse function's graph?

1. An inverse function exchanges the domain and range of the original function. Therefore, \(D_g = R_f = \{x \in \mathbb{R} \mid x > 0\}\) and \(R_g = D_f = \mathbb{R}\). 2. Points on inverse functions have their coordinates reversed. Since \((0, 1)\) lies on the graph of \(f\), \((1, 0)\) lies on the graph of \(g\). Thus, \(g\) has an x-intercept at \(x = 1\).

1. \(D_g = \{x \in \mathbb{R} \mid x > 0\}\); \(R_g = \mathbb{R}\) 2. \(P'(1, 0)\); the x-intercept is \(x = 1\).

- How do a point's coordinates change when reflected across \(y=x\)? - How will an increasing set of points look after their coordinates are swapped? - What happens to the horizontal line \(y=0\) when the coordinate roles switch? - As the exponential outputs approach \(0\), what happens to the corresponding inverse inputs?

1. Reflection across \(y=x\) swaps the coordinates of every point. Since \(f\) is increasing and one-to-one, its inverse \(g(x)=\log_a(x)\) is also increasing. 2. Under the reflection, the horizontal asymptote \(y=0\) of \(f\) becomes the vertical asymptote \(x=0\) of \(g\). As \(x\to0^+\), \(g(x)\to-\infty\).

1. \(g\) is increasing because it is the inverse of the increasing one-to-one function \(f\). 2. Vertical asymptote: \(x=0\), the y-axis.

- Evaluate the logarithmic term when \(x = 1\). - Rewrite an equation of the form \(\log_b(u) = v\) in exponential form. - Recall the definition of a function in terms of inputs and outputs.

1. Substitute \(A(1, 5)\): \(5 = \log_b(1) + c\). Since \(\log_b(1) = 0\) for every valid base, \(c = 5\). 2. Substitute \(B(16, 9)\) and \(c = 5\): \(9 = \log_b(16) + 5\), so \(\log_b(16) = 4\). 3. Rewrite in exponential form: \(b^4 = 16\). Because a logarithm base must be positive, \(b = 2\). Thus, \(f(x) = \log_2(x) + 5\). 4. At \(x = 1\), the logarithmic term is always \(0\), so the function value equals \(c\). This makes \(c\) equal to the y-coordinate of any point with x-coordinate \(1\). 5. A function cannot assign two different outputs to the same input. The points \((2, 3)\) and \((2, 4)\) would require both \(f(2) = 3\) and \(f(2) = 4\), so no such function exists.

a) \(b = 2\), \(c = 5\), and \(f(x) = \log_2(x) + 5\) b) \(\log_b(1) = 0\), so \(f(1) = c\) for every valid base. c) No. A function cannot have two different outputs for the same input.

- Rewrite each logarithmic equation in exponential form. - When the base is unknown, solve the resulting power equation and apply the base restrictions. - Express reciprocals and roots as powers of the relevant base. - Fractional exponents can make root relationships easier to recognize.

1. a) Rewrite in exponential form: \(x = 2^5 = 32\). 2. b) \(x^4 = 81\). A logarithm base must be positive, so \(x = 3\). 3. c) Since \(\frac{1}{64} = 4^{-3}\), \(y = -3\). 4. d) Since \(3 = 9^{1/2}\), \(x = \frac{1}{2}\). 5. e) Since \(\sqrt{1000} = \sqrt{10^3} = 10^{3/2}\), \(y = \frac{3}{2}\).

a) \(x = 32\) b) \(x = 3\) c) \(y = -3\) d) \(x = \frac{1}{2}\) e) \(y = \frac{3}{2}\)

- Rewrite each logarithmic equation as an exponential equation. - Recall the special rules for exponents of \(0\) and \(1\). - Apply the restrictions on a valid logarithm base. - Interpret a negative exponent as a reciprocal.

1. Rewrite each equation in exponential form. 2. a) \(a^3 = 64\), so \(a = 4\). 3. b) \(a^1 = 11\), so \(a = 11\). 4. c) \(a^0 = 5\). Every valid logarithm base satisfies \(a^0 = 1\), so this would require \(1 = 5\). There is no solution. 5. d) \(a^{-2} = \frac{1}{25}\), so \(a^2 = 25\). Although the power equation has solutions \(a = \pm 5\), a logarithm base must be positive. Therefore, \(a = 5\).

a) \(a = 4\) b) \(a = 11\) c) No solution d) \(a = 5\)

- Rewrite each logarithmic statement as an exponential equation. - How can you undo a fractional exponent? - What is the value of a valid base raised to the zeroth power, and does that give a unique base? - Remember that a logarithm base must be positive and cannot equal \(1\).

1. Use \(\log_b(x) = y\) if and only if \(b^y = x\). 2. a) \(b^3 = 1000\), so \(b = \sqrt[3]{1000} = 10\). 3. b) \(b^{1/3} = 2\). Cubing both sides gives \(b = 8\). 4. c) \(b^{-1} = 0.1 = \frac{1}{10}\), so \(\frac{1}{b} = \frac{1}{10}\) and \(b = 10\). 5. d) \(b^0 = 1\). This is true for every valid logarithm base, so every \(b > 0\) with \(b \ne 1\) is a solution.

a) \(b = 10\) b) \(b = 8\) c) \(b = 10\) d) Every \(b > 0\) with \(b \ne 1\)

- Rewrite a logarithmic equation as an exponential equation. - Which positive number cubed equals \(125\)? - Express the reciprocal as a power with a negative exponent. - Use the inverse relationship between logarithms and exponentials.

1. a) Substitute the point into the function: \(\log_a(125) = 3\). In exponential form, \(a^3 = 125\), so \(a = 5\). 2. b) Now \(f(x) = \log_5(x)\). Since \(\frac{1}{25} = 5^{-2}\), \(f\left(\frac{1}{25}\right) = -2\). 3. c) Solve \(\log_5(x) = 5\) by rewriting it as \(x = 5^5\). Therefore, \(x = 3125\).

a) \(a = 5\) b) \(f\left(\frac{1}{25}\right) = -2\) c) \(x = 3125\)

- Interpret the negative exponent after rewriting the logarithmic equation in exponential form. - Remember that a logarithm base must be positive. - To test a point, substitute its coordinates into the function equation.

1. a) Substitute the coordinates of \(Q\): \(-2 = \log_b\left(\frac{1}{9}\right)\). 2. Rewrite in exponential form: \(b^{-2} = \frac{1}{9}\). 3. This gives \(\frac{1}{b^2} = \frac{1}{9}\), so \(b^2 = 9\). A logarithm base must be positive, so \(b = 3\). 4. b) Test the coordinates of \(R\) in \(y = \log_3(x)\). Since \(\log_3(27) = 3\) because \(3^3 = 27\), the point lies on the graph.

a) \(b = 3\) b) Yes. \(R(27, 3)\) lies on the graph because \(\log_3(27) = 3\).

- Use the definition of a logarithm to relate the base, exponent, and argument. - Which first power equals \(10\)? - Recall the end behavior of a logarithmic function with a base greater than \(1\). - Compare a logarithm with base \(b\) to one with reciprocal base \(\frac{1}{b}\).

1. Substitute \(P(10, 1)\): \(1 = \log_b(10)\). In exponential form, \(b^1 = 10\), so \(b = 10\). 2. Because the base \(10\) is greater than \(1\), \(h(x) \to -\infty\) as \(x \to 0^+\), and \(h(x) \to \infty\) as \(x \to \infty\). 3. The base of \(k\) is \(\frac{1}{10}\). Since \(\left(\frac{1}{10}\right)^{-2} = 10^2 = 100\), \(k(100) = -2\).

1. \(b = 10\) 2. \(h(x) \to -\infty\) as \(x \to 0^+\), and \(h(x) \to \infty\) as \(x \to \infty\). 3. \(k(100) = -2\)

- Which exponential function is the inverse of \(\ln(x)\)? - How can you use the calculator to narrow the interval first to hundredths, then thousandths? - What is the halfway point between consecutive thousandth values, and why does it matter for rounding?

1. Narrow the interval using \(e^{1.60}\approx4.9530\) and \(e^{1.61}\approx5.0028\), giving \(1.60<\ln(5)<1.61\). 2. Testing thousandths gives \(e^{1.609}\approx4.9978\) and \(e^{1.610}\approx5.0028\). Thus \(1.609<\ln(5)<1.610\). 3. The rounding midpoint is \(1.6095\), and \(e^{1.6095}\approx5.0003>5\). Therefore \(\ln(5)<1.6095\), so \(\ln(5)\approx1.609\) to three decimal places.

a) \(e^{1.609}\approx4.9978<5<5.0028\approx e^{1.610}\), so \(1.609<\ln(5)<1.610\). b) \(e^{1.6095}\approx5.0003>5\), so \(\ln(5)\approx1.609\) to three decimal places.

- Rewrite \(y=\ln(x)\) in exponential form. - For an arbitrary target output \(S\), identify an input that produces it. - Apply the exponential function to solve each equation.

1. The equation \(y=\ln(x)\) is equivalent to \(x=e^y\). 2. For every positive bound \(S\), the input \(x=e^S\) satisfies \(\ln(x)=S\). Therefore, logarithmic outputs can exceed every fixed bound, so \(f\) is unbounded above. 3. If \(\ln(x)=15\), then \(x=e^{15}\approx3{,}269{,}017.37\). 4. If \(\ln(x)=35\), then \(x=e^{35}\approx1.586\times10^{15}\).

The function is unbounded above because \(\ln(e^S)=S\) for every \(S>0\). For \(f(x)=15\), \(x=e^{15}\approx3{,}269{,}017.37\). For \(f(x)=35\), \(x=e^{35}\approx1.586\times10^{15}\).

- Compare the graphs' x-intercepts and vertical asymptotes. - Look at how the graphs change between \(x=1\) and \(x=2\). - Which graph shifts left, and which reflects across the x-axis?

1. Graph \(f\) goes through \((1, 0)\), \((2, 1)\), and \((4, 2)\), matching equation 1. 2. Graph \(g\) goes through \((1, 0)\), \((2, 2)\), and \((4, 4)\), matching equation 2, a vertical stretch of \(f\). 3. Graph \(h\) crosses the x-axis at \((-2, 0)\) and has vertical asymptote \(x=-3\), matching equation 3. 4. Graph \(k\) crosses at \((1, 0)\) and falls as \(x\) increases, matching equation 4, a reflection across the x-axis. 5. Equation 5 is unused.

\(f\): 1; \(g\): 2; \(h\): 3; \(k\): 4. Equation 5 is unused.

- Pay attention to the axis units: \(x\) represents thousands of bacteria. - Look for points whose coordinates lie exactly on grid intersections. - For \(y=\log_b(x)\), the value of \(x\) when \(y=1\) equals the base \(b\).

1. a) The graph contains the points \((2, 1)\) and \((8, 3)\). Since \(x\) is measured in thousands, there are \(2000\) bacteria after \(1\) hour and \(8000\) bacteria after \(3\) hours. 2. b) Substitute \((2, 1)\) into \(y=\log_b(x)\): \(1=\log_b(2)\). Therefore, \(b^1=2\), so \(b=2\). 3. The function is \(y=\log_2(x)\). 4. c) For \(16{,}000\) bacteria, \(x=16\). Then \(y=\log_2(16)=4\), so the culture reaches \(16{,}000\) bacteria after \(4\) hours.

a) After \(1\) hour: \(2000\) bacteria; after \(3\) hours: \(8000\) bacteria b) \(y=\log_2(x)\) c) \(4\) hours

- Use a graph point whose x-coordinate is \(-1\) or \(1\) to identify the base. - What operation reverses exponentiation with a fixed base? - Express both \(8\) and \(0.5\) as powers of \(2\).

1. a) The graph contains the point \((-1, 2)\). Substituting gives \(a^{-1} = 2\), so \(\frac{1}{a} = 2\) and \(a = 0.5\). Thus, \(f(x) = (0.5)^x\). 2. b) The inverse of an exponential function with base \(a\) is the logarithmic function with the same base. Therefore, \(f^{-1}(x) = \log_{0.5}(x)\). 3. c) To find \(f^{-1}(8)\), solve \((0.5)^y = 8\). Since \(0.5 = 2^{-1}\) and \(8 = 2^3\), \(2^{-y} = 2^3\). Hence, \(-y = 3\) and \(y = -3\).

a) \(a = 0.5\) b) \(f^{-1}(x) = \log_{0.5}(x)\) c) \(f^{-1}(8) = -3\)

- Which inputs and outputs can you read from the marked points? - For \(b^x+c\), what does the graph's height at input \(0\) tell you about \(c\)? - Which features exchange roles for a function and its inverse? - How can the original graph's y-intercept help locate the inverse's x-intercept?

1. The marked point \(P=(0, -1)\) gives \(1+c=-1\), so \(c=-2\). 2. The marked point \(Q=(1, 2)\) gives \(b-2=2\), so \(b=4\) and \(f(x)=4^x-2\). 3. The range of \(f\) is \((-2,\infty)\), so the domain of \(f^{-1}\) is \((-2,\infty)\). The range of \(f^{-1}\) is all real numbers. 4. Reflecting the horizontal asymptote \(y=-2\) across \(y=x\) gives the inverse's vertical asymptote \(x=-2\). The point \(P\) becomes \((-1, 0)\), the inverse's x-intercept. 5. Reversing \(y=4^x-2\) gives \(f^{-1}(x)=\log_4(x+2)\).

\(b=4\), \(c=-2\) Rule: \(f^{-1}(x)=\log_4(x+2)\) Domain: \((-2,\infty)\) Range: \((-\infty,\infty)\) Vertical asymptote: \(x=-2\) x-intercept: \((-1, 0)\)

- First isolate the expression containing the exponent. - Which function reverses raising \(3\) to a power? - Keep the logarithmic expression exact until the final rounding step.

1. Divide by \(4\): \(3^{0.5x}=\frac{19}{4}\). 2. Apply the inverse base-3 logarithm: \(0.5x=\log_3\left(\frac{19}{4}\right)\). 3. Multiply by \(2\): \(x=2\log_3\left(\frac{19}{4}\right)\). 4. Numerically, \(x\approx2.84\).

\(x=2\log_3\left(\frac{19}{4}\right)\approx2.84\)

- Set the exponential model equal to the target number of cells. - Isolate the exponential before changing to logarithmic form. - Read the approximate crossing height on the graph to check your result.

1. Set \(200\cdot2^{t/3}=950\). Dividing by \(200\) gives \(2^{t/3}=4.75\). 2. In logarithmic form, \(\frac{t}{3}=\log_{2}(4.75)\), so \(t=3\log_{2}(4.75)\). 3. The value is \(t\approx6.7\) hours, consistent with the graph crossing the height \(950\) between \(6\) and \(7\) hours.

a) \(\frac{t}{3}=\log_{2}(4.75)\) b) \(t=3\log_{2}(4.75)\approx6.7\,\text{hours}\)

- For what values of \(x\) is \(16 - x^2\) positive? - How does a decreasing function compare outputs when the inputs increase? - Rewrite the logarithmic equation in exponential form. - When the unknown is the logarithm's base, remember all restrictions on a valid base.

1. a) The logarithm's argument must be positive: \(16 - x^2 > 0\). Thus, \(x^2 < 16\), so \(-4 < x < 4\). 2. b) The order of the arguments is reversed only when the logarithmic function is decreasing. Therefore, \(0 < a < 1\). 3. c) Since \(4 > 1\) but \(\log_a(4) = -2 < 0\), the logarithmic function must be decreasing, so \(0 < a < 1\). Algebraically, \(a^{-2} = 4\), so \(a^2 = \frac{1}{4}\). The base restriction \(a > 0\) gives \(a = \frac{1}{2}\). 4. d) Rewrite the equation as \(x^4 = 625\). The real solutions of this power equation are \(x = \pm 5\), but a logarithm base must be positive and cannot equal \(1\). Therefore, the only valid solution is \(x = 5\).

a) \((-4, 4)\) b) \(0 < a < 1\) c) \(0 < a < 1\); in fact, \(a = \frac{1}{2}\) d) \(x = 5\)

- What must be true of the expression inside the logarithm? - Express the number on the right as a logarithm with the same base. - How does a base between \(0\) and \(1\) affect comparisons of logarithmic values? - Check which parts of the inequality solution satisfy the domain.

1. The logarithm requires \(x^2-8>0\), so \(x<-\sqrt{8}\) or \(x>\sqrt{8}\). 2. Since \(\left(\frac{1}{3}\right)^{-2}=9\), rewrite the inequality as \(\log_{\frac{1}{3}}(x^2-8)>\log_{\frac{1}{3}}(9)\). 3. The logarithm decreases because its base is between \(0\) and \(1\). Reversing the inequality gives \(x^2-8<9\), or \(-\sqrt{17}<x<\sqrt{17}\). 4. Combining this with the domain gives \(( -\sqrt{17},-\sqrt{8})\cup(\sqrt{8},\sqrt{17})\).

Domain: \(x<-\sqrt{8}\) or \(x>\sqrt{8}\). Solution: \((-\sqrt{17},-\sqrt{8})\cup(\sqrt{8},\sqrt{17})\). Since the base is less than \(1\), the logarithm is decreasing, so the inequality reverses when comparing its arguments.

- Rewrite the logarithmic equation in exponential form. - Clear the fractional exponent by raising both sides to a power. - Compare a basic number property of the two resulting integer powers. - Unique prime factorization gives another way to see the contradiction.

1. Assume for contradiction that \(\log_2(5) = \frac{m}{n}\), where \(m\) and \(n\) are positive integers. 2. Rewrite the logarithmic equation in exponential form: \(2^{m/n} = 5\). 3. Raise both sides to the \(n\)th power: \(2^m = 5^n\). 4. Because \(m \ge 1\), \(2^m\) is even. Because \(n \ge 1\), \(5^n\) is odd. 5. An even integer cannot equal an odd integer, so \(2^m = 5^n\) is impossible. This contradiction shows that \(\log_2(5)\) is irrational.

Assuming \(\log_2(5) = \frac{m}{n}\) leads to \(2^m = 5^n\), but the left side is even and the right side is odd. Therefore, \(\log_2(5)\) is irrational.

- Factor the base \(10\) into primes. - Clear the fractional exponent first. - Compare the prime factors that must occur on each side. - Use the uniqueness of prime factorization to obtain a contradiction.

1. Assume for contradiction that \(\log_{10}(2) = \frac{p}{q}\), where \(p\) and \(q\) are positive integers. 2. Rewrite in exponential form: \(10^{p/q} = 2\). Raising both sides to the \(q\)th power gives \(10^p = 2^q\). 3. Factor the left side: \(10^p = (2 \cdot 5)^p = 2^p5^p\). 4. Thus, \(2^p5^p = 2^q\). The left side has a prime factor of \(5\), while the right side has only the prime factor \(2\). 5. This contradicts the uniqueness of prime factorization. Therefore, \(\log_{10}(2)\) is irrational.

The rationality assumption leads to \(10^p = 2^q\), or \(2^p5^p = 2^q\). The two sides cannot have the same prime factorization, so \(\log_{10}(2)\) is irrational.