Free Math Worksheets

- 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) \(x = 8\)

- 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) \(\frac{1}{3}\)

- Rewrite each root as a rational exponent. - Express the argument as a power of the logarithm base. - A reciprocal can be written using a negative exponent. - You may also use the power property of logarithms after rewriting the argument.

1. a) Rewrite the argument as a power of \(2\): \(\sqrt[3]{16} = (2^4)^{1/3} = 2^{4/3}\). Therefore, the value is \(\frac{4}{3}\). 2. b) Rewrite the argument as a power of \(5\): \(\sqrt[4]{125} = (5^3)^{1/4} = 5^{3/4}\). Therefore, the value is \(\frac{3}{4}\). 3. c) Rewrite the argument as a power of \(3\): \(\frac{1}{\sqrt{27}} = 27^{-1/2} = (3^3)^{-1/2} = 3^{-3/2}\). Therefore, the value is \(-\frac{3}{2}\). 4. d) Since \(0.1 = 10^{-1}\), \(\sqrt[5]{0.1} = (10^{-1})^{1/5} = 10^{-1/5}\). Therefore, the value 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 \(3\)

- Look for consecutive integer powers of the base that bracket the argument. - For arguments between \(0\) and \(1\), consider negative powers. - Use \(b^x = a\) if and only if \(x = \log_b(a)\).

1. a) \(5^{-3} = 0.008\) and \(5^{-2} = 0.04\). Since \(0.008 < 0.03 < 0.04\), \(-3 < \log_5(0.03) < -2\). 2. b) \(8^2 = 64\) and \(8^3 = 512\). Since \(64 < 100 < 512\), \(2 < \log_8(100) < 3\). 3. c) \(2^1 = 2\) and \(2^2 = 4\). Since \(2 < 2.5 < 4\), \(1 < \log_2(2.5) < 2\). 4. d) \(10^{-3} = 0.001\) and \(10^{-2} = 0.01\). Since \(0.001 < 0.002 < 0.01\), \(-3 < \log_{10}(0.002) < -2\).

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) \(b = 5\)

- 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.

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

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

- 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. a) \(\log_2(64) = 6\) because \(2^6 = 64\). Also, \(\sqrt{10} = 10^{1/2}\), so \(\log_{10}(\sqrt{10}) = \frac{1}{2}\). 2. b) \(\log_b(b^5) = 5\). The logarithm asks for the exponent on \(b\) that produces \(b^5\), and that exponent is \(5\). 3. c) 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.

- Determine whether \(\log_{0.5}(10)\) is positive or negative. - Recall what happens to an inequality when both sides are multiplied by a negative number. - Consider whether a logarithmic function with base between \(0\) and \(1\) is increasing or decreasing.

1. Since \(0 < 0.5 < 1\) and \(10 > 1\), \(\log_{0.5}(10)\) is negative. In fact, \(\log_{0.5}(10) = \frac{\log(10)}{\log(0.5)} \approx -3.32\). 2. Multiplying \(3 > 2\) by the negative number \(\log_{0.5}(10)\) reverses the inequality: \(3\log_{0.5}(10) < 2\log_{0.5}(10)\). 3. Numerically, the two sides are approximately \(-9.97\) and \(-6.64\), so the left side is smaller. 4. Also, \(y = \log_{0.5}(x)\) is decreasing. Because \(1000 > 100\), it follows that \(\log_{0.5}(1000) < \log_{0.5}(100)\).

The first inequality is incorrect. Since \(\log_{0.5}(10) < 0\), multiplying by it reverses the inequality. Therefore, \(\log_{0.5}(1000) < \log_{0.5}(100)\), which also follows because a logarithmic function with base between \(0\) and \(1\) is decreasing.

- Rewrite each root as a rational exponent. - Express both logarithm arguments as powers of \(a\). - Use \(\log_a(a^r) = r\). - Divide fractions by multiplying by the reciprocal.

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

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

- 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: \(x + 3x = 4x\).

\(4x\)

- 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) \(b = 5\)

- Determine which exponents on each base produce \(2\), \(4\), and \(8\). - Rewrite \(0.5\) as a power of \(2\). - Use the change-of-base formula with base \(2\). - Interpret the transformation from \((x, y)\) to \((x, -y)\).

1. a) \(f(2) = 1\), \(f(4) = 2\), and \(f(8) = 3\). 2. Since \(0.5^{-1} = 2\), \(0.5^{-2} = 4\), and \(0.5^{-3} = 8\), \(g(2) = -1\), \(g(4) = -2\), and \(g(8) = -3\). 3. The values of \(g\) are the opposites of the corresponding values of \(f\). 4. b) By change of base, \(g(x) = \frac{\log_2(x)}{\log_2(0.5)}\). Since \(0.5 = 2^{-1}\), \(\log_2(0.5) = -1\). Thus, \(g(x) = -\log_2(x) = -f(x)\). 5. c) Replacing every output \(y\) with \(-y\) reflects a graph across the x-axis. Therefore, the graph of \(g\) is the reflection of the graph of \(f\) across the x-axis.

a) \(f(2)=1\), \(f(4)=2\), \(f(8)=3\); \(g(2)=-1\), \(g(4)=-2\), \(g(8)=-3\). The values are opposites. b) \(g(x) = \frac{\log_2(x)}{\log_2(2^{-1})} = -f(x)\) c) Reflection across the x-axis

- Write the transformed function as \(g(x) = kf(x)\). - Use the change-of-base formula with common logarithms. - Identify the constant multiplying \(\log(x)\).

1. A vertical stretch gives \(g(x) = kf(x)\), so \(\log_5(x) = k\log(x)\). 2. By change of base, \(\log_5(x) = \frac{\log(x)}{\log(5)}\). 3. Therefore, \(k = \frac{1}{\log(5)}\). 4. Since \(\log(5) \approx 0.69897\), \(k \approx 1.43068 \approx 1.43\).

\(k = \frac{1}{\log(5)} \approx 1.43\)

- 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) \(x = 3125\)

- 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 \(f(8) = 3\).

- 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. \(k(100) = -2\)

- Use the monotonicity of \(e^x\) to translate exponential inequalities into bounds for \(\ln(5)\). - Narrow the interval by testing hundredths and then thousandths. - Compare with the midpoint needed for rounding to three decimal places.

1. Test hundredths: \(e^{1.60}\approx4.9530\) and \(e^{1.61}\approx5.0028\), so \(1.60<\ln(5)<1.61\). 2. Test thousandths: \(e^{1.609}\approx4.9978\) and \(e^{1.610}\approx5.0028\), so \(1.609<\ln(5)<1.610\). 3. To decide the rounding, compare with the midpoint \(1.6095\). Since \(e^{1.6095}\approx5.0003>5\), \(\ln(5)<1.6095\). 4. Therefore, \(\ln(5)\approx1.609\) to three decimal places.

\(\ln(5)\approx1.609\)

- 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 x-intercepts and vertical asymptotes of the graphs. - Check the output at \(x = 2\) or \(x = 4\). - Look for horizontal shifts, reflections, and vertical stretches. - An expression of the form \(\log_2(x + c)\) shifts the base graph horizontally.

1. Graph \(f\) passes through \((1, 0)\), \((2, 1)\), and \((4, 2)\), so it matches \(y = \log_2(x)\), equation 1. 2. Graph \(g\) passes through \((1, 0)\), \((2, 2)\), and \((4, 4)\). Its outputs are twice those of \(f\), so it matches \(y = 2\log_2(x)\), equation 2. 3. Graph \(h\) has x-intercept \((-2, 0)\) and vertical asymptote \(x = -3\). It is the base graph shifted \(3\) units left, so it matches \(y = \log_2(x + 3)\), equation 3. 4. Graph \(k\) passes through \((1, 0)\), \((2, -1)\), and \((4, -2)\). It is the reflection of \(f\) across the x-axis, so it matches \(y = -\log_2(x)\), equation 4. 5. Equation 5 is not used.

\(f \rightarrow 1\) \(g \rightarrow 2\) \(h \rightarrow 3\) \(k \rightarrow 4\) 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

- 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) \(x = 5\)

- Require the logarithm argument to be positive. - Rewrite \(-2\) as a logarithm with base \(\frac{1}{3}\). - Determine whether the logarithmic function is increasing or decreasing. - Intersect the inequality solution with the domain.

1. The logarithm requires \(x^2-8 > 0\), so \(x^2 > 8\). Thus, \(x < -\sqrt{8}\) or \(x > \sqrt{8}\). 2. Rewrite the right side as \(-2 = \log_{\frac{1}{3}}(9)\). 3. Because \(0 < \frac{1}{3} < 1\), the logarithmic function is decreasing. Therefore, \(\log_{\frac{1}{3}}(x^2-8) > \log_{\frac{1}{3}}(9)\) implies \(x^2-8 < 9\). 4. Solve \(x^2 < 17\): \(-\sqrt{17} < x < \sqrt{17}\). 5. Intersect this interval with the domain. The solution is \((-\sqrt{17}, -\sqrt{8}) \cup (\sqrt{8}, \sqrt{17})\).

\((-\sqrt{17}, -\sqrt{8}) \cup (\sqrt{8}, \sqrt{17})\), approximately \((-4.12, -2.83) \cup (2.83, 4.12)\)

- 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.