Free Math Worksheets

- 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) \(x = 16\)

- Rewrite each logarithmic equation in exponential form. - Recall what a logarithm with base \(b\) tells you about an exponent. - Remember the value of a nonzero number raised to the zeroth power. - After removing the logarithm, solve the remaining linear equation for \(x\).

1. a) Rewrite the equation in exponential form: \(x = 2^6 = 64\). 2. b) Rewrite in exponential form: \(2x + 15 = 5^3 = 125\). Then \(2x = 110\), so \(x = 55\). 3. c) Rewrite in exponential form: \(x - 4 = 3^0 = 1\). Therefore, \(x = 5\). 4. d) Rewrite in exponential form: \(0.1x = 10^2 = 100\). Multiplying by \(10\) gives \(x = 1000\).

a) \(x = 64\) b) \(x = 55\) c) \(x = 5\) d) \(x = 1000\)

- Rewrite each logarithmic equation in exponential form. - Which positive number cubed equals \(125\)? - Interpret a negative exponent as a reciprocal. - Rewrite \(9\) as a power of \(3\). - Apply all restrictions on a logarithm base.

Use \(\log_b(u) = v\) if and only if \(b^v = u\). 1. a) \(x^3 = 125\), so \(x = 5\). 2. b) \(x = 2^{-4} = \frac{1}{16} = 0.0625\). 3. c) \(9^x = 3\). Since \(9 = 3^2\), \(3^{2x} = 3^1\), so \(2x = 1\) and \(x = \frac{1}{2}\). 4. d) \(x^{-2} = \frac{1}{16}\), so \(x^2 = 16\). The power equation gives \(x = \pm 4\), but a logarithm base must be positive. Therefore, \(x = 4\).

a) \(x = 5\) b) \(x = \frac{1}{16}\) c) \(x = \frac{1}{2}\) d) \(x = 4\)

- Use the relationship between logarithmic and exponential forms. - Rewrite each equation so the logarithm is removed. - Identify the base, exponent, and power carefully. - Rewrite a radical base using a rational exponent.

1. Use \(\log_b(u) = v\) if and only if \(b^v = u\). 2. For 1), \(x = 4^{-2} = \frac{1}{16}\). 3. For 2), \(x^3 = 64\), so \(x = 4\). 4. For 3), \(x = (\sqrt{5})^4 = (5^{1/2})^4 = 25\). 5. For 4), \(\left(\frac{1}{3}\right)^x = 27\). Since \(\frac{1}{3} = 3^{-1}\) and \(27 = 3^3\), \(3^{-x} = 3^3\), so \(x = -3\).

1) \(x = \frac{1}{16}\) 2) \(x = 4\) 3) \(x = 25\) 4) \(x = -3\)

- Convert each logarithmic equation to exponential form. - Identify the unknown base in the corresponding power equation. - Ask which positive number raised to the given exponent produces the argument. - Apply the restrictions on a valid logarithm base.

1. Use \(\log_x(u) = v\) if and only if \(x^v = u\). 2. a) \(x^3 = 125\), so \(x = 5\). 3. b) \(x^5 = 32\), so \(x = 2\). 4. c) \(x^{-2} = 0.01 = \frac{1}{100}\), so \(x^2 = 100\). The power equation gives \(x = \pm 10\), but a logarithm base must be positive. Therefore, \(x = 10\).

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

- Convert each logarithmic equation to exponential form. - Which positive number cubed equals \(125\)? - Relate reciprocals to negative exponents. - Remember that a logarithm base must be positive.

1. Use \(\log_b(u) = v\) if and only if \(b^v = u\). 2. For 1), \(x^3 = 125\), so \(x = 5\). 3. For 2), \(2^x = \frac{1}{32} = 2^{-5}\), so \(x = -5\). 4. For 3), \(x^{-2} = 0.25 = \frac{1}{4}\), so \(x^2 = 4\). A logarithm base must be positive, so \(x = 2\). 5. For 4), \(9^x = 3\). Since \(3 = 9^{1/2}\), \(x = \frac{1}{2}\).

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

- Use the definition of a logarithm as the inverse of exponentiation. - Rewrite \(\log_a(u) = v\) as \(a^v = u\). - Apply all restrictions when the unknown is the logarithm base. - Rewrite decimals as fractions or powers of \(10\) when helpful.

1. For 1), \(x = 2^{-3} = \frac{1}{8}\). 2. For 2), \(x^4 = 81\). The power equation has real solutions \(x = \pm 3\), but a logarithm base must be positive. Therefore, \(x = 3\). 3. For 3), \((0.1)^x = 100\). Rewrite as \((10^{-1})^x = 10^2\), so \(-x = 2\) and \(x = -2\). 4. For 4), \(x = \left(\frac{1}{2}\right)^5 = \frac{1}{32}\).

1) \(x = \frac{1}{8}\) 2) \(x = 3\) 3) \(x = -2\) 4) \(x = \frac{1}{32}\)

- Rewrite each argument using \(3\) and a power of \(10\). - Rewrite a square root as a power of \(\frac{1}{2}\). - Use the product, quotient, and power properties. - Recall that \(\log(10^n) = n\).

1. \(\log(300) = \log(3 \cdot 100) = \log(3) + 2 \approx 2.4771\). 2. \(\log(0.003) = \log\left(\frac{3}{1000}\right) = \log(3) - 3 \approx -2.5229\). 3. \(\log(81) = \log(3^4) = 4\log(3) \approx 1.9084\). 4. \(\log(\sqrt{3}) = \log(3^{1/2}) = \frac{1}{2}\log(3) \approx 0.23855 \approx 0.2386\).

1) \(\approx 2.4771\) 2) \(\approx -2.5229\) 3) \(\approx 1.9084\) 4) \(\approx 0.2386\)

- Express each argument as \(5\) times a power of \(10\). - Use the product property of logarithms. - Recall that \(\log_{10}(10^k) = k\).

1. Write each argument as \(5 \cdot 10^k\) and use \(\log_{10}(5 \cdot 10^k) = \log_{10}(5) + k\). 2. a) \(\log_{10}(5000) = \log_{10}(5 \cdot 10^3) \approx 0.6990 + 3 = 3.6990\). 3. b) \(\log_{10}(0.5) = \log_{10}(5 \cdot 10^{-1}) \approx 0.6990 - 1 = -0.3010\). 4. c) \(\log_{10}(0.005) = \log_{10}(5 \cdot 10^{-3}) \approx 0.6990 - 3 = -2.3010\).

a) \(\approx 3.6990\) b) \(\approx -0.3010\) c) \(\approx -2.3010\)

- Use the power property to move coefficients into exponents. - A sum of logarithms becomes the logarithm of a product. - A difference of logarithms becomes the logarithm of a quotient. - Rewrite an exponent of \(\frac{1}{2}\) as a square root.

1. For part 1, apply the power property: \(\log(x) = \log(a^2) + \log(b^3)\). 2. Apply the product property: \(\log(x) = \log(a^2b^3)\), so \(x = a^2b^3\). 3. For part 2, use \(\frac{1}{2}\log(a) = \log(\sqrt{a})\). 4. Apply the quotient property: \(\log(x) = \log\left(\frac{\sqrt{a}}{b}\right)\), so \(x = \frac{\sqrt{a}}{b}\).

1) \(x = a^2b^3\) 2) \(x = \frac{\sqrt{a}}{b}\)

- Expand products and quotients into sums and differences of logarithms. - Rewrite roots and reciprocals using exponents. - Recall that \(\log_b(b) = 1\). - Combine the coefficients of \(\log_b(x)\).

1. Expand the first term: \(\log_b(b\sqrt{x}) = 1 + \frac{1}{2}\log_b(x)\). 2. Expand the second term: \(\log_b\left(\frac{b^2}{x}\right) = 2 - \log_b(x)\). 3. Rewrite the third term: \(-\frac{1}{2}\log_b(x^{-1}) = \frac{1}{2}\log_b(x)\). 4. Combine like terms: \(A = 1 + \frac{1}{2}\log_b(x) + 2 - \log_b(x) + \frac{1}{2}\log_b(x) = 3\).

\(A = 3\)

- Express \(12\) and \(1.5\) using \(2\) and \(3\). - Use the change-of-base formula when the logarithm base is not \(10\). - Apply the power property when an argument contains an exponent.

1. a) Since \(12 = 2^2 \cdot 3\), \(\log(12) = 2\log(2) + \log(3) \approx 2(0.3010) + 0.4771 = 1.0791 \approx 1.079\). 2. b) By change of base, \(\log_2(3) = \frac{\log(3)}{\log(2)} \approx \frac{0.4771}{0.3010} \approx 1.585\). 3. c) \(\log_3(1.5) = \log_3\left(\frac{3}{2}\right) = 1-\log_3(2) = 1-\frac{\log(2)}{\log(3)} \approx 0.369\).

a) \(\log(12) \approx 1.079\) b) \(\log_2(3) \approx 1.585\) c) \(\log_3(1.5) \approx 0.369\)

- Start with the outermost operation in the logarithm argument. - Rewrite the cube root as a rational exponent. - Evaluate the logarithm of the power of \(10\). - Use the product, quotient, and power properties.

1. Apply the quotient property: \(x = \log(100a^4) - \log(\sqrt[3]{b})\). 2. Apply the product property: \(x = \log(100) + \log(a^4) - \log(b^{1/3})\). 3. Since \(\log(100) = 2\), apply the power property to obtain \(x = 2 + 4\log(a) - \frac{1}{3}\log(b)\).

\(x = 2 + 4\log(a) - \frac{1}{3}\log(b)\)

- Rewrite the logarithmic equation in exponential form. - Express the decimal exponent as a fraction if helpful. - Solve the resulting linear equation. - Check that the logarithm argument is positive.

1. Rewrite in exponential form: \(2x-1 = 4^{2.5}\). 2. Evaluate the power: \(4^{2.5} = 4^{5/2} = 32\). 3. Solve the linear equation: \(2x-1 = 32\), so \(2x = 33\) and \(x = \frac{33}{2}\). 4. Check the domain: \(2\left(\frac{33}{2}\right)-1 = 32 > 0\), so the solution is valid.

\(\left\{\frac{33}{2}\right\}\)

- Convert each logarithmic equation to an exponential equation. - Identify the positive number that produces the argument when raised to the given exponent. - Apply the restrictions on a logarithm base. - Rewrite a root using a fractional exponent.

1. Rewrite each logarithmic equation in exponential form. 2. a) \(b^2 = 144\). A logarithm base must be positive, so \(b = 12\). 3. b) \(b^{-3} = \frac{1}{27}\), so \(b^3 = 27\) and \(b = 3\). 4. c) \(b^3 = 216\), so \(b = 6\). 5. d) \(b^{1/3} = 5^{1/3}\). Since both bases are positive, \(b = 5\).

a) \(b = 12\) b) \(b = 3\) c) \(b = 6\) d) \(b = 5\)

- Use the change-of-base formula for a logarithm with an arbitrary base. - You may use either common logarithms or natural logarithms. - Check that your result lies between \(2\) and \(3\), since \(6^2 < 45 < 6^3\).

1. Apply the change-of-base formula: \(\log_6(45) = \frac{\log(45)}{\log(6)}\). 2. Using a calculator, \(\log(45) \approx 1.653213\) and \(\log(6) \approx 0.778151\). 3. Divide: \(\frac{1.653213}{0.778151} \approx 2.124539\). 4. Rounded to four decimal places, the value is \(2.1245\).

\(\approx 2.1245\)

- Use the change-of-base formula with common or natural logarithms. - Compare \(12\) with consecutive integer powers of \(5\). - Use the definition of a logarithm to translate the power comparison.

1. Use the change-of-base formula: \(\log_5(12) = \frac{\ln(12)}{\ln(5)}\). 2. The quotient is approximately \(1.543959\), which rounds to \(1.544\). 3. Since \(5^1 = 5\) and \(5^2 = 25\), \(5 < 12 < 25\). 4. Therefore, \(1 < \log_5(12) < 2\).

\(\log_5(12) \approx 1.544\), and it lies between \(1\) and \(2\).

- Use the change-of-base formula for both logarithms. - Use the same standard logarithm base in both calculations. - Round only after computing each quotient.

1. By change of base, \(\log_2(10) = \frac{\ln(10)}{\ln(2)} \approx 3.321928 \approx 3.322\). 2. Also, \(\log_3(30) = \frac{\ln(30)}{\ln(3)} \approx 3.095903 \approx 3.096\). 3. Since \(3.322 > 3.096\), \(\log_2(10) > \log_3(30)\).

\(\log_2(10) \approx 3.322\) and \(\log_3(30) \approx 3.096\), so \(\log_2(10)\) is greater.

- Express \(16\), \(4\), and \(\sqrt{2}\) as powers of \(2\). - Use the quotient property for a reciprocal. - Use the product property to separate factors. - Apply the power property to each power of \(2\).

1. Apply the quotient property to the first term: \(\log_2\left(\frac{1}{16}\right) = \log_2(1) - \log_2(16) = 0 - 4 = -4\). 2. Apply the product property to the second term: \(\log_2(4\sqrt{2}) = \log_2(4) + \log_2(\sqrt{2})\). 3. Apply the power property using \(4 = 2^2\) and \(\sqrt{2} = 2^{1/2}\): the second term is \(2 + \frac{1}{2} = \frac{5}{2}\). 4. Therefore, \(T = -4 + \frac{5}{2} = -\frac{3}{2}\).

\(T = -\frac{3}{2}\)

- Express each number as \(6\) times a power of \(10\). - Use the product property of logarithms. - Recall that \(\log(10^k) = k\).

1. a) \(\log(60) = \log(6 \cdot 10) = \log(6) + 1 \approx 1.778\). 2. \(\log(6000) = \log(6 \cdot 10^3) = \log(6) + 3 \approx 3.778\). 3. \(\log(0.06) = \log(6 \cdot 10^{-2}) = \log(6) - 2 \approx -1.222\). 4. b) For any positive \(x\), \(\log(x \cdot 10^k) = \log(x) + \log(10^k) = \log(x) + k\). Multiplying by \(10^k\) adds \(k\) to the common logarithm.

a) \(\log(60) \approx 1.778\), \(\log(6000) \approx 3.778\), and \(\log(0.06) \approx -1.222\) b) The logarithm increases by \(k\): \(\log(x \cdot 10^k) = \log(x) + k\).

- Express each argument as a product, quotient, or power involving \(3\) and \(5\). - Recall that \(\log_5(5) = 1\). - Rewrite \(0.6\) as a fraction. - Rewrite the square root as a power of \(\frac{1}{2}\).

1. \(\log_5(9) = \log_5(3^2) = 2\log_5(3) \approx 1.366\). 2. \(\log_5(15) = \log_5(3 \cdot 5) = \log_5(3) + 1 \approx 1.683\). 3. \(\log_5(0.6) = \log_5\left(\frac{3}{5}\right) = \log_5(3) - 1 \approx -0.317\). 4. \(\log_5(\sqrt{3}) = \frac{1}{2}\log_5(3) \approx 0.3415\).

1. \(\approx 1.366\) 2. \(\approx 1.683\) 3. \(\approx -0.317\) 4. \(\approx 0.3415\)

- Express each argument using \(2\) and the base \(5\). - Use the product, quotient, and power properties. - Recall that \(\log_5(5) = 1\). - Convert terminating decimals to fractions when helpful.

1. a) \(\log_5(4) = \log_5(2^2) = 2\log_5(2) \approx 0.86\). 2. b) \(\log_5(10) = \log_5(2 \cdot 5) = \log_5(2) + 1 \approx 1.43\). 3. c) \(\log_5(0.4) = \log_5\left(\frac{2}{5}\right) = \log_5(2) - 1 \approx -0.57\). 4. d) \(\log_5(\sqrt{2}) = \frac{1}{2}\log_5(2) \approx 0.215\). 5. e) \(\log_5(2.5) = \log_5\left(\frac{5}{2}\right) = 1-\log_5(2) \approx 0.57\).

a) \(\approx 0.86\) b) \(\approx 1.43\) c) \(\approx -0.57\) d) \(\approx 0.215\) e) \(\approx 0.57\)

- Express \(15\) as a product of \(3\) and \(5\). - Use the product property of logarithms. - Add the lower bounds to get a lower bound for the sum, and add the upper bounds to get an upper bound.

1. Use the product property: \(\log(15) = \log(3) + \log(5)\). 2. Add the lower bounds: \(0.4771 + 0.6989 = 1.1760\). 3. Add the upper bounds: \(0.4772 + 0.6990 = 1.1762\). 4. Therefore, \(1.1760 < \log(15) < 1.1762\).

\(1.1760 < \log(15) < 1.1762\)

- Identify the products and quotients inside each logarithm. - A product becomes a sum of logarithms. - A quotient becomes a difference of logarithms. - Keep parentheses around a denominator product until it is fully expanded. - Simplify \(\log_k(k)\).

1. a) \(\log_k(kmn) = \log_k(k) + \log_k(m) + \log_k(n) = 1 + \log_k(m) + \log_k(n)\). 2. b) \(\log_k\left(\frac{5}{kx}\right) = \log_k(5) - \log_k(kx) = \log_k(5) - 1 - \log_k(x)\). 3. c) \(\log_k\left(\frac{ab}{c}\right) = \log_k(a) + \log_k(b) - \log_k(c)\).

a) \(1 + \log_k(m) + \log_k(n)\) b) \(\log_k(5) - 1 - \log_k(x)\) c) \(\log_k(a) + \log_k(b) - \log_k(c)\)

- Apply the product and quotient properties in reverse. - A sum of logarithms becomes the logarithm of a product. - A difference of logarithms becomes the logarithm of a quotient. - Simplify numerical factors before writing the final logarithm. - A sum such as \(x+1\) cannot be split by a logarithm property.

1. a) Use the product and quotient properties: \(\log_y(a) + \log_y(b) - \log_y(c) = \log_y\left(\frac{ab}{c}\right)\). 2. b) First simplify \(\frac{15}{3} = 5\): \(\log_y(15) - \log_y(3) - \log_y(z) = \log_y\left(\frac{5}{z}\right)\). 3. c) Apply the quotient property: \(\log_y(x+1) - \log_y(x) = \log_y\left(\frac{x+1}{x}\right)\).

a) \(\log_y\left(\frac{ab}{c}\right)\) b) \(\log_y\left(\frac{5}{z}\right)\) c) \(\log_y\left(\frac{x+1}{x}\right)\)

- Identify the exponent inside each logarithm. - Rewrite a root as a rational exponent. - Rewrite a reciprocal using a negative exponent. - Move the exponent in front of the logarithm.

1. a) \(\log_b(x^9) = 9\log_b(x)\). 2. b) \(\log_b(y^{-4}) = -4\log_b(y)\). 3. c) Since \(\sqrt[3]{z} = z^{1/3}\), \(\log_b\left(\sqrt[3]{z}\right) = \frac{1}{3}\log_b(z)\). 4. d) Since \(\frac{1}{a^5} = a^{-5}\), \(\log_b\left(\frac{1}{a^5}\right) = -5\log_b(a)\). 5. e) Since \(0.75 = \frac{3}{4}\), \(\log_b(w^{0.75}) = \frac{3}{4}\log_b(w)\).

a) \(9\log_b(x)\) b) \(-4\log_b(y)\) c) \(\frac{1}{3}\log_b(z)\) d) \(-5\log_b(a)\) e) \(\frac{3}{4}\log_b(w)\)

- Rewrite each logarithm argument as one power. - Use \(\frac{1}{x^r} = x^{-r}\) and \(\sqrt[n]{x^m} = x^{m/n}\). - Simplify the argument before applying the power property. - Move the final exponent in front of the logarithm.

1. a) \(\sqrt[5]{x^2} = x^{2/5}\), so the expression is \(\frac{2}{5}\log_k(x)\). 2. b) \(\frac{1}{\sqrt{y}} = y^{-1/2}\), so the expression is \(-\frac{1}{2}\log_k(y)\). 3. c) \(\frac{1}{z^{-3}} = z^3\), so the expression is \(3\log_k(z)\). 4. d) \(\sqrt[n]{a^m} = a^{m/n}\), so the expression is \(\frac{m}{n}\log_k(a)\).

a) \(\frac{2}{5}\log_k(x)\) b) \(-\frac{1}{2}\log_k(y)\) c) \(3\log_k(z)\) d) \(\frac{m}{n}\log_k(a)\)

- Use the power property to move each coefficient into an exponent. - Combine added logarithms with the product property. - Combine the subtraction with the quotient property. - Rewrite an exponent of \(\frac{1}{3}\) as a cube root.

1. Apply the power property: \(5\log_b(p) = \log_b(p^5)\), \(\frac{1}{3}\log_b(q) = \log_b(\sqrt[3]{q})\), and \(2\log_b(r) = \log_b(r^2)\). 2. Use the product property on the first two terms: \(\log_b(p^5) + \log_b(\sqrt[3]{q}) = \log_b(p^5\sqrt[3]{q})\). 3. Use the quotient property: \(\log_b(p^5\sqrt[3]{q}) - \log_b(r^2) = \log_b\left(\frac{p^5\sqrt[3]{q}}{r^2}\right)\).

\(\log_b\left(\frac{p^5\sqrt[3]{q}}{r^2}\right)\)

- Identify the quotient and products inside the argument. - Apply the quotient property first. - Separate all factors with the product property. - Move exponents and root indices in front using the power property.

1. Apply the quotient property: \(\log_a(5x^2\sqrt[3]{y}) - \log_a(z^4)\). 2. Apply the product property: \(\log_a(5) + \log_a(x^2) + \log_a(\sqrt[3]{y}) - \log_a(z^4)\). 3. Apply the power property: \(\log_a(5) + 2\log_a(x) + \frac{1}{3}\log_a(y) - 4\log_a(z)\).

\(\log_a(5) + 2\log_a(x) + \frac{1}{3}\log_a(y) - 4\log_a(z)\)

- Combine logarithms with the same base. - Addition of logarithms corresponds to multiplication of arguments. - Subtraction of logarithms corresponds to division of arguments. - Express the final argument as a power of \(10\).

1. Use the product property: \(\log(120) + \log(25) = \log(3000)\). 2. Use the quotient property: \(\log(3000) - \log(3) = \log(1000)\). 3. Since \(1000 = 10^3\), \(\log(1000) = 3\).

\(3\)

- Combine the sum of logarithms into the logarithm of a product. - Simplify the fractions before multiplying. - Express the final argument as a power of \(2\).

1. Use the product property: \(\log_2\left(\frac{14}{5} \cdot \frac{10}{7} \cdot 8\right)\). 2. Simplify the product: \(\frac{14}{5} \cdot \frac{10}{7} \cdot 8 = 2 \cdot 2 \cdot 8 = 32\). 3. Since \(32 = 2^5\), the value is \(5\).

\(5\)

- Evaluate \(\sqrt[3]{64}\) and the two base-\(2\) logarithms. - Rewrite a root as a rational exponent. - Apply the power property with \(k = \frac{1}{n}\).

1. The left side is \(\log_2\left(\sqrt[3]{64}\right) = \log_2(4) = 2\). 2. The right side is \(\frac{1}{3}\log_2(64) = \frac{1}{3}(6) = 2\). 3. Both sides equal \(2\), so the identity holds for the given values. 4. In general, \(\log_a\left(\sqrt[n]{u}\right) = \log_a(u^{1/n})\). Applying the power property gives \(\frac{1}{n}\log_a(u)\).

For the given values, both sides equal \(2\). In general, \(\log_a\left(\sqrt[n]{u}\right) = \log_a(u^{1/n}) = \frac{1}{n}\log_a(u)\).

- You do not need to determine \(b\) or \(x\). - Use the power property for the exponent in part a. - Rewrite the fifth root as a power of \(\frac{1}{5}\).

1. a) By the power property, \(\log_b(x^3) = 3\log_b(x) = 3 \cdot 1.2 = 3.6\). 2. b) Since \(\sqrt[5]{x} = x^{1/5}\), \(\log_b\left(\sqrt[5]{x}\right) = \frac{1}{5}\log_b(x) = \frac{1}{5} \cdot 1.2 = 0.24\).

a) \(3.6\) b) \(0.24\)

- 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) \(b = 8\)

- Use the change-of-base formula for each nonstandard base. - Round only after evaluating the quotient. - Check whether an argument is an exact integer power of its base.

1. Use \(\log_b(a) = \frac{\log(a)}{\log(b)}\). 2. a) \(\log_7(100) = \frac{\log(100)}{\log(7)} \approx 2.367\). 3. b) \(\log_{0.5}(10) = \frac{\log(10)}{\log(0.5)} \approx -3.322\). 4. c) \(\log_{12}(144) = 2\) because \(12^2 = 144\).

a) \(\log_7(100) \approx 2.367\) b) \(\log_{0.5}(10) \approx -3.322\) c) \(\log_{12}(144) = 2\)

- Use the change-of-base formula to rewrite each logarithm with base \(10\). - The calculator key labeled “log” gives the common logarithm. - Enter the logarithm of the argument in the numerator and the logarithm of the base in the denominator. - Check whether any values can also be recognized exactly from powers.

1. Use \(\log_b(a)=\frac{\log(a)}{\log(b)}\). 2. a) \(\log_8(32)=\frac{\log(32)}{\log(8)}=\frac{5}{3}\approx1.667\). 3. b) \(\log_6(15)=\frac{\log(15)}{\log(6)}\approx1.511\). 4. c) \(\log_{0.5}(4)=\frac{\log(4)}{\log(0.5)}=-2\). 5. d) \(\log_{12}(100)=\frac{\log(100)}{\log(12)}\approx1.853\).

a) \(\frac{5}{3}\approx1.667\) b) \(\approx1.511\) c) \(-2\) d) \(\approx1.853\)

- Logarithmic and exponential functions with the same base are inverses. - When the logarithm's argument is an expression, first rewrite the equation in exponential form. - The definition of a logarithm also applies when the base is the unknown. - Interpret a negative exponent as a reciprocal.

1. a) \(\log_3(9) = 2\) and \(\log_5(125) = 3\), so \(2 \cdot 2 - 3 = 1\). 2. b) Rewrite in exponential form: \(x + 10 = 2^6 = 64\). Therefore, \(x = 54\). 3. c) Logarithmic and exponential functions with base \(10\) are inverses, so \(x + 5 = 8\). Thus, \(x = 3\). 4. d) Rewrite in exponential form: \(x = 4^{-2} = \frac{1}{16} = 0.0625\). 5. e) Rewrite in exponential form: \(x^4 = 81\). The real solutions of the power equation are \(x = \pm 3\), but a logarithm base must be positive. Therefore, \(x = 3\).

a) \(1\) b) \(x = 54\) c) \(x = 3\) d) \(x = \frac{1}{16}\) e) \(x = 3\)

- When the base is unknown, rewrite the logarithmic equation in exponential form and solve the resulting power equation. - Recall what a negative exponent means. - Use the fact that logarithmic and exponential functions with the same base are inverses. - Write \(1.5\) as \(\frac{3}{2}\) to evaluate the power without a calculator.

1. a) Rewrite the equation as \(x^4 = 81\). A logarithm base must be positive and cannot equal \(1\), so the valid solution is \(x = 3\). 2. b) Rewrite the equation as \(x = 4^{-1}\). Therefore, \(x = \frac{1}{4} = 0.25\). 3. c) Since logarithms and exponential functions with the same base are inverses, \(3x - 1 = 8\). Thus, \(3x = 9\) and \(x = 3\). 4. d) Rewrite the equation as \(3x = 9^{1.5} = 9^{3/2} = 27\). Therefore, \(x = 9\).

a) \(x = 3\) b) \(x = 0.25\) c) \(x = 3\) d) \(x = 9\)

- State the restrictions on a logarithm base before accepting solutions. - Treat a multi-term base or argument as a single quantity when converting to exponential form. - Rewrite a terminating decimal as a fraction when useful. - Interpret a negative exponent as a reciprocal.

1. For 1), rewrite the equation as \(x - 4 = 2^5 = 32\). Therefore, \(x = 36\). 2. For 2), \((x + 2)^2 = 49\), so \(x + 2 = 7\) or \(x + 2 = -7\). A logarithm base must be positive and cannot equal \(1\), so only \(x + 2 = 7\) is valid. Thus, \(x = 5\). 3. For 3), \(x = (0.2)^{-3} = \left(\frac{1}{5}\right)^{-3} = 125\). 4. For 4), \(x^{-2} = \frac{1}{100}\), so \(x^2 = 100\). The power equation gives \(x = \pm 10\), but the base must be positive. Therefore, \(x = 10\).

1) \(x = 36\) 2) \(x = 5\) 3) \(x = 125\) 4) \(x = 10\)

- Rewrite each logarithmic statement in exponential form. - Express the decimal exponent as a fraction. - To solve \(x^{r} = c\), apply the reciprocal exponent when the quantities are valid. - Which exponent produces \(1\) from every valid base?

1. a) Rewrite the equation as \(2x - 1 = 3^2 = 9\). Then \(2x = 10\), so \(x = 5\). 2. b) Rewrite the equation as \(x^{0.75} = 27\). Since \(0.75 = \frac{3}{4}\), raise both sides to the power \(\frac{4}{3}\): \(x = 27^{4/3} = (\sqrt[3]{27})^4 = 3^4 = 81\). 3. c) Let \(b > 0\) and \(b \ne 1\). The equation \(\log_b(1) = c\) is equivalent to \(b^c = 1\). Since \(b^0 = 1\) and \(b^x\) is one-to-one for every valid base, the unique exponent is \(c = 0\). Therefore, \(\log_b(1) = 0\).

a) \(x = 5\) b) \(x = 81\) c) For every valid base \(b\), \(b^0 = 1\), and the exponential function is one-to-one. Therefore, \(\log_b(1) = 0\).

- Rewrite fractions and radicals as powers. - Try to express both sides using compatible exponents. - Apply the reciprocal exponent to isolate the base. - Use exponent rules carefully and remember the base restrictions.

1. Use \(\log_x(u) = v\) if and only if \(x^v = u\). 2. a) \(x^{-7} = \frac{1}{128}\), so \(x^7 = 128\). Therefore, \(x = 2\). 3. b) \(x^{3/2} = 2\sqrt{2}\). Since \(2\sqrt{2} = 2^{3/2}\) and a logarithm base is positive, \(x = 2\). 4. c) \(x^{3/2} = 27\). Raise both sides to the power \(\frac{2}{3}\): \(x = 27^{2/3} = (\sqrt[3]{27})^2 = 9\).

a) \(x = 2\) b) \(x = 2\) c) \(x = 9\)

- Rewrite the decimal exponent as a fraction. - Interpret a negative exponent when the base is itself a fraction. - Express the square root as a rational exponent. - Compare the three exact values after solving.

1. For (1), \(x^{1.5} = 8\). Since \(1.5 = \frac{3}{2}\), raise both sides to the power \(\frac{2}{3}\): \(x = 8^{2/3} = 4\). 2. For (2), \(x = (0.5)^{-3} = \left(\frac{1}{2}\right)^{-3} = 8\). 3. For (3), \(10^x = \sqrt{10} = 10^{1/2}\), so \(x = \frac{1}{2}\). 4. The three values are \(4\), \(8\), and \(\frac{1}{2}\). The least value is \(\frac{1}{2}\).

(1) \(x = 4\) (2) \(x = 8\) (3) \(x = \frac{1}{2}\) The least value is \(\frac{1}{2}\).

- Rewrite roots using rational exponents. - Solving an equation involving \(x^2\) may produce two real values; check both in the logarithm. - Why must a logarithm return a unique exponent? - Review the restrictions on a logarithm base.

1. Rewrite the equation as \(x^{1/6} = \sqrt{2} = 2^{1/2}\). Raise both sides to the sixth power: \(x = (2^{1/2})^6 = 2^3 = 8\). 2. Rewrite the equation as \(x^2 + 2 = 3^3 = 27\). Then \(x^2 = 25\), so \(x = -5\) or \(x = 5\). Both values make the logarithm's argument equal to \(27\), so both are valid. 3. A logarithm base must be positive and cannot equal \(1\). If the base were \(1\), the equation \(1^y = 1\) would be true for every real \(y\), so the logarithm would not have a unique value. Therefore, \(\log_1(1)\) is undefined.

1) \(x = 8\) 2) \(x \in \{-5, 5\}\) 3) \(\log_1(1)\) is undefined because base \(1\) is not allowed; \(1^y = 1\) does not determine a unique exponent.

- Apply the restrictions on both the base and the argument of a real logarithm. - Which values can never be logarithm bases? - Is a real logarithm defined for every possible argument? - When the logarithm is defined, rewrite the equation in exponential form.

For a defined logarithm, the base must satisfy \(x > 0\) and \(x \ne 1\), and the argument must satisfy \(a > 0\). When these conditions hold, \(\log_x(a) = a\) is equivalent to \(x^a = a\). 1. a) \(x^3 = 3\), so \(x = \sqrt[3]{3}\). This is a valid base. 2. b) \(x^1 = 1\), so \(x = 1\). Since base \(1\) is not allowed, there is no solution. 3. c) \(x^{1/4} = \frac{1}{4}\), so \(x = \left(\frac{1}{4}\right)^4 = \frac{1}{256}\). This is a valid base. 4. d) The argument \(-1\) is not positive, so \(\log_x(-1)\) is undefined for every real logarithm base. There is no solution.

a) \(x = \sqrt[3]{3}\) b) No solution c) \(x = \frac{1}{256}\) d) No solution

- Use the power property to move a coefficient into an exponent. - Use the product property for sums of logarithms. - Use the quotient property for differences of logarithms. - Combine each side into a single logarithm before comparing arguments.

1. a) Apply the power property: \(\log(x) = \log(a^2) + \log(b^3) - \log(c)\). 2. Apply the product and quotient properties: \(\log(x) = \log\left(\frac{a^2b^3}{c}\right)\). 3. Because logarithmic functions are one-to-one, \(x = \frac{a^2b^3}{c}\). 4. b) Apply the power property: \(\log(x) = \log(u^{1/2}) + \log(u+v)\). 5. Apply the product property: \(\log(x) = \log\left(\sqrt{u}(u+v)\right)\). 6. Therefore, \(x = (u+v)\sqrt{u}\).

a) \(x = \frac{a^2b^3}{c}\) b) \(x = (u+v)\sqrt{u}\)

- Combine the logarithms inside the parentheses first. - Use the power property for each coefficient. - A difference of logarithms becomes the logarithm of a quotient. - Rewrite the entire right side as one logarithm.

1. Use the product property inside the parentheses: \(\log(a) + \log(b) = \log(ab)\). 2. Apply the power property: \(\frac{1}{2}\log(ab) = \log\left(\sqrt{ab}\right)\) and \(3\log(c) = \log(c^3)\). 3. Use the quotient property: \(\log(x) = \log\left(\frac{\sqrt{ab}}{c^3}\right)\). 4. Therefore, \(x = \frac{\sqrt{ab}}{c^3}\).

\(x = \frac{\sqrt{ab}}{c^3}\)

- Express \(1\) as a common logarithm. - Use the power property to move a coefficient into an exponent. - Combine sums and differences of logarithms into one logarithm.

1. Use the power property: \(2\log(6) = \log(36)\). 2. Write \(1 = \log(10)\). 3. Use the quotient property: \(\log(36) - \log(9) = \log(4)\). 4. Use the product property: \(\log(4) + \log(10) = \log(40)\). 5. Therefore, \(x = 40\).

\(x = 40\)

- Use the power property to move coefficients into exponents. - Combine sums and differences of logarithms into one logarithm. - Substitute the changed values directly into your expression for \(x\). - Compare the simplified new expression with the original one.

1. a) Apply the power property: \(2\log(a) = \log(a^2)\) and \(\frac{1}{2}\log(b) = \log(\sqrt{b})\). 2. Combine the logarithms: \(\log(x) = \log\left(\frac{a^2\sqrt{b}}{c}\right)\). 3. Therefore, \(x = \frac{a^2\sqrt{b}}{c}\). 4. b) Substitute \(2a\) for \(a\) and \(4c\) for \(c\): \(x_{\text{new}} = \frac{(2a)^2\sqrt{b}}{4c} = \frac{4a^2\sqrt{b}}{4c} = \frac{a^2\sqrt{b}}{c}\). 5. Thus, \(x\) does not change.

a) \(x = \frac{a^2\sqrt{b}}{c}\) b) \(x\) remains unchanged because the factor of \(4\) from squaring \(2a\) cancels the factor of \(4\) in the denominator.

- Use the power property to move a coefficient into an exponent. - Combine a sum of logarithms with the product property. - Combine a difference of logarithms with the quotient property. - An exponent of \(\frac{1}{2}\) represents a square root.

1. For part 1, apply the power property: \(\log(x) = \log(a^4) + \log(b^2)\). 2. Apply the product property: \(\log(x) = \log(a^4b^2)\), so \(x = a^4b^2\). 3. For part 2, apply the power and quotient properties: \(\log(x) = \frac{1}{2}\log\left(\frac{c}{d^4}\right)\). 4. Move the coefficient into the exponent: \(\log(x) = \log\left(\left(\frac{c}{d^4}\right)^{1/2}\right)\). 5. Therefore, \(x = \frac{\sqrt{c}}{d^2}\).

1) \(x = a^4b^2\) 2) \(x = \frac{\sqrt{c}}{d^2}\)

- Factor \(a^2-9\) as a difference of squares. - Use the quotient property to combine the first two logarithms. - Express \(1\) as a common logarithm. - Check the domain restrictions from the original equation.

1. Use the quotient property: \(\log(x) = \log\left(\frac{a^2-9}{a+3}\right) + 1\). 2. Factor and simplify: \(\frac{a^2-9}{a+3} = \frac{(a-3)(a+3)}{a+3} = a-3\). 3. Write \(1 = \log(10)\), then use the product property: \(\log(x) = \log(10(a-3))\). 4. Therefore, \(x = 10(a-3) = 10a-30\). 5. The original logarithms require \(a^2-9 > 0\) and \(a+3 > 0\), so \(a > 3\).

\(x = 10a-30\), for \(a > 3\)

- Simplify the expression inside the parentheses first. - Use the power property to move coefficients into exponents. - Combine the final difference with the quotient property. - Rewrite an exponent of \(\frac{1}{3}\) as a cube root.

1. Combine the logarithms inside the parentheses: \(\log(b) + 2\log(c) = \log(bc^2)\). 2. Apply the power property: \(\frac{1}{3}\log(bc^2) = \log\left(\sqrt[3]{bc^2}\right)\) and \(2\log(a) = \log(a^2)\). 3. Apply the quotient property: \(\log(x) = \log\left(\frac{a^2}{\sqrt[3]{bc^2}}\right)\). 4. Therefore, \(x = \frac{a^2}{\sqrt[3]{bc^2}}\).

\(x = \frac{a^2}{\sqrt[3]{bc^2}}\)

- Distribute the coefficient outside the parentheses first. - Use the power property to move coefficients into exponents. - Subtracted logarithms produce factors in the denominator. - Rewrite an exponent of \(\frac{1}{2}\) as a square root.

1. Distribute the coefficient: \(\log(x) = 3\log(a) - \frac{1}{2}\log(b) - 2\log(c)\). 2. Apply the power property: \(\log(x) = \log(a^3) - \log(\sqrt{b}) - \log(c^2)\). 3. Apply the quotient property: \(\log(x) = \log\left(\frac{a^3}{c^2\sqrt{b}}\right)\). 4. Therefore, \(x = \frac{a^3}{c^2\sqrt{b}}\).

\(x = \frac{a^3}{c^2\sqrt{b}}\)

- Express each argument using \(2\), \(3\), and powers of \(10\). - Use the product property for multiplication and the quotient property for division. - Write \(5\) as \(\frac{10}{2}\). - Use the power property when an argument contains a square.

1. \(\log(6) = \log(2 \cdot 3) = \log(2) + \log(3) \approx 0.7781\). 2. \(\log(1.5) = \log\left(\frac{3}{2}\right) = \log(3) - \log(2) \approx 0.1761\). 3. \(\log(5) = \log\left(\frac{10}{2}\right) = 1 - \log(2) \approx 0.6990\). 4. \(\log(18) = \log(2 \cdot 3^2) = \log(2) + 2\log(3) \approx 1.2552\).

1) \(\approx 0.7781\) 2) \(\approx 0.1761\) 3) \(\approx 0.6990\) 4) \(\approx 1.2552\)

- Express each argument using \(2\), \(3\), and powers of \(10\). - Factor \(12\) into powers of \(2\) and \(3\). - Write \(4.5\) as a quotient involving \(3^2\) and \(2\). - Write \(5\) as \(\frac{10}{2}\).

1. a) Since \(12 = 2^2 \cdot 3\), \(\log(12) = 2\log(2) + \log(3) \approx 2(0.301) + 0.477 = 1.079\). 2. b) Since \(4.5 = \frac{3^2}{2}\), \(\log(4.5) = 2\log(3) - \log(2) \approx 2(0.477) - 0.301 = 0.653\). 3. c) Since \(5 = \frac{10}{2}\), \(\log(5) = 1 - \log(2) \approx 1 - 0.301 = 0.699\).

a) \(\log(12) \approx 1.079\) b) \(\log(4.5) \approx 0.653\) c) \(\log(5) \approx 0.699\)

- Express each argument using only \(2\), \(3\), and powers of \(10\). - Use addition for products and subtraction for quotients inside logarithms. - Recall that \(\log(10) = 1\). - Convert a terminating decimal to a fraction when helpful.

1. Since \(18 = 2 \cdot 3^2\), \(\log(18) = a + 2b \approx 0.301 + 2(0.477) = 1.255\). 2. Since \(5 = \frac{10}{2}\), \(\log(5) = 1-a \approx 1-0.301 = 0.699\). 3. Since \(0.75 = \frac{3}{2^2}\), \(\log(0.75) = b-2a \approx 0.477-2(0.301) = -0.125\).

1. \(\log(18) \approx 1.255\) 2. \(\log(5) \approx 0.699\) 3. \(\log(0.75) \approx -0.125\)

- Compare the defining properties of arithmetic and geometric sequences. - Use the quotient property on the difference of two logarithms. - Find the ratio \(\frac{b_{n+1}}{b_n}\). - You may also simplify the formula for \(a_n\) before taking a difference.

1. a) The first three terms of \(b_n\) are \(b_1 = 3\), \(b_2 = 6\), and \(b_3 = 12\). 2. Therefore, \(a_1 = \log(3)\), \(a_2 = \log(6)\), and \(a_3 = \log(12)\). 3. b) Compute the difference: \(a_{n+1}-a_n = \log(b_{n+1})-\log(b_n) = \log\left(\frac{b_{n+1}}{b_n}\right)\). 4. Since \(b_n\) is geometric with common ratio \(2\), \(\frac{b_{n+1}}{b_n} = 2\). 5. Thus, \(a_{n+1}-a_n = \log(2)\), which is independent of \(n\). Therefore, \(a_n\) is arithmetic with common difference \(d = \log(2)\).

a) \(a_1 = \log(3)\), \(a_2 = \log(6)\), \(a_3 = \log(12)\) b) \(d = \log(2)\), so \(a_n\) is arithmetic.

- Separate the quotient and product using logarithm properties. - Rewrite the square root as a power of \(\frac{1}{2}\). - Simplify the logarithm whose argument equals its base. - Substitute the given expressions only after expanding.

1. Apply the quotient and product properties: \(L = \log_a(u^2) + \log_a(\sqrt{v}) - \log_a(a)\). 2. Apply the power property and use \(\log_a(a) = 1\): \(L = 2\log_a(u) + \frac{1}{2}\log_a(v) - 1\). 3. Substitute \(x = \log_a(u)\) and \(y = \log_a(v)\): \(L = 2x + \frac{1}{2}y - 1\).

\(L = 2x + \frac{1}{2}y - 1\)

- Determine how many factors of \(10\) take \(a\) to \(b\). - Use the product property of logarithms. - Relate a decimal-place shift to multiplication by a power of \(10\).

1. a) \(\log(2.5) \approx 0.398\) and \(\log(250) \approx 2.398\). 2. b) The difference is \(2.398-0.398 = 2\). 3. c) Since \(250 = 2.5 \cdot 10^2\), \(\log(250) = \log(2.5) + \log(10^2) = \log(2.5) + 2\). 4. Therefore, \(\log(b)-\log(a) = 2\) exactly.

a) \(\log(2.5) \approx 0.398\); \(\log(250) \approx 2.398\) b) \(2\) c) Because \(b = a \cdot 10^2\), \(\log(b) = \log(a) + 2\), so the difference is exactly \(2\).

- Use the product property for multiplication. - Use the quotient property for division. - Rewrite a fourth root as a power of \(\frac{1}{4}\). - Express \(820\) using \(8.2\) and a power of \(10\).

1. a) Use the product property: \(\log(3.6 \cdot 8.2) \approx 0.5563 + 0.9138 = 1.4701\). 2. b) Use the quotient property: \(\log\left(\frac{8.2}{3.6}\right) \approx 0.9138 - 0.5563 = 0.3575\). 3. c) Use the power property: \(\log\left((3.6)^{1/4}\right) \approx \frac{1}{4}(0.5563) = 0.139075\). 4. d) Since \(820 = 8.2 \cdot 10^2\), \(\log(820) \approx 0.9138 + 2 = 2.9138\).

a) \(\approx 1.4701\) b) \(\approx 0.3575\) c) \(\approx 0.139075\) d) \(\approx 2.9138\)

- Use the product and quotient properties for parts a and b. - Use the power property for the exponent and the cube root. - Express \(25\) as \(2.5 \cdot 10\). - Recall that \(\log(10) = 1\).

1. a) Use the product property: \(\log(2.5 \cdot 5.4) \approx 0.3979 + 0.7324 = 1.1303\). 2. b) Use the quotient property: \(\log\left(\frac{5.4}{2.5}\right) \approx 0.7324 - 0.3979 = 0.3345\). 3. c) Use the power property: \(\log(2.5^3) \approx 3(0.3979) = 1.1937\). 4. d) Since \(25 = 2.5 \cdot 10\), \(\log\left(\sqrt[3]{25}\right) = \frac{1}{3}\left(\log(2.5)+1\right) \approx \frac{1.3979}{3} \approx 0.465967\).

a) \(\approx 1.1303\) b) \(\approx 0.3345\) c) \(\approx 1.1937\) d) \(\approx 0.465967\)

- Express each argument as a product, quotient, or power involving \(2\), \(3\), and \(10\). - Write \(5\) as \(\frac{10}{2}\). - Apply the product, quotient, and power properties systematically. - Recall that \(\log_{10}(10) = 1\).

1. a) \(\log_{10}(6) = \log_{10}(2) + \log_{10}(3) \approx 0.778\). 2. b) \(\log_{10}(1.5) = \log_{10}(3) - \log_{10}(2) \approx 0.176\). 3. c) \(\log_{10}(12) = 2\log_{10}(2) + \log_{10}(3) \approx 1.079\). 4. d) \(\log_{10}(5) = 1-\log_{10}(2) \approx 0.699\). 5. e) Since \(180 = 2 \cdot 3^2 \cdot 10\), \(\log_{10}(180) = \log_{10}(2) + 2\log_{10}(3) + 1 \approx 2.255\).

a) \(\approx 0.778\) b) \(\approx 0.176\) c) \(\approx 1.079\) d) \(\approx 0.699\) e) \(\approx 2.255\)

- Express \(1.5\) as \(\frac{3}{2}\). - Use the quotient property of logarithms. - Determine which choices of bounds make a difference smallest and largest.

1. Write \(1.5 = \frac{3}{2}\), so \(\log(1.5) = \log(3) - \log(2)\). 2. The smallest possible difference comes from the lower bound for \(\log(3)\) and the upper bound for \(\log(2)\): \(0.4771-0.3011 = 0.1760\). 3. The largest possible difference comes from the upper bound for \(\log(3)\) and the lower bound for \(\log(2)\): \(0.4772-0.3010 = 0.1762\). 4. Therefore, \(0.1760 < \log(1.5) < 0.1762\).

\(0.1760 < \log(1.5) < 0.1762\)

- Rewrite the numerical constant as a logarithm with base \(a\). - Apply the power property to the coefficients first. - Combine added logarithms with the product property. - Combine the subtraction with the quotient property.

1. Apply the power property: \(\frac{1}{2}\log_a(x^2+y^2) = \log_a\left(\sqrt{x^2+y^2}\right)\) and \(2\log_a(x) = \log_a(x^2)\). 2. Rewrite the constant as \(3 = \log_a(a^3)\). 3. Use the product and quotient properties: \(\log_a\left(\sqrt{x^2+y^2}\right) + \log_a(x^2) - \log_a(a^3) = \log_a\left(\frac{x^2\sqrt{x^2+y^2}}{a^3}\right)\).

\(\log_a\left(\frac{x^2\sqrt{x^2+y^2}}{a^3}\right)\)

- Factor \(u^2-v^2\) as a difference of squares. - Apply the quotient and power properties first. - Expand the logarithm of the factored product. - Combine like logarithm terms.

1. Apply the quotient and power properties: \(2\log_k(u-v) - \frac{1}{2}\log_k(u^2-v^2)\). 2. Factor the difference of squares: \(u^2-v^2 = (u-v)(u+v)\). 3. Apply the product property: \(2\log_k(u-v) - \frac{1}{2}\left(\log_k(u-v) + \log_k(u+v)\right)\). 4. Combine like terms: \(\frac{3}{2}\log_k(u-v) - \frac{1}{2}\log_k(u+v)\).

\(\frac{3}{2}\log_k(u-v) - \frac{1}{2}\log_k(u+v)\)

- Rewrite both logarithms using the same standard base. - Look for a common factor that cancels. - Apply the proved identity directly in part b. - Express \(9\) as a power of \(3\).

1. a) Apply change of base: \(\log_a(b)\log_b(c) = \frac{\log(b)}{\log(a)} \cdot \frac{\log(c)}{\log(b)}\). 2. Cancel \(\log(b)\): \(\frac{\log(c)}{\log(a)} = \log_a(c)\). This proves the identity. 3. b) Apply the identity: \(\log_3(5)\log_5(9) = \log_3(9)\). 4. Since \(3^2 = 9\), the value is \(2\).

a) \(\frac{\log(b)}{\log(a)} \cdot \frac{\log(c)}{\log(b)} = \frac{\log(c)}{\log(a)} = \log_a(c)\) b) \(2\)

- Start with the general change-of-base formula. - Use natural logarithms as the auxiliary base. - Recognize \(81\) as a power of \(3\). - Consider whether the auxiliary base changes the value of the original logarithm.

1. a) Apply change of base with natural logarithms: \(\log_3(x) = \frac{\ln(x)}{\ln(3)}\). 2. b) \(\log_3(10) = \frac{\ln(10)}{\ln(3)} \approx 2.0959\). 3. Since \(81 = 3^4\), \(\log_3(81) = 4\). 4. c) The change-of-base formula works with any valid auxiliary base. Both quotients equal \(\log_3(10)\), so they must be equal to each other.

a) \(\log_3(x) = \frac{\ln(x)}{\ln(3)}\) b) \(\log_3(10) \approx 2.0959\); \(\log_3(81) = 4\) c) Both quotients equal \(\log_3(10)\) by the change-of-base formula.

- Look for a logarithm base that matches the next logarithm’s argument. - Apply the change-of-base identity one pair at a time. - Express \(81\) as a power of \(3\).

1. Use \(\log_a(b)\log_b(c) = \log_a(c)\): \(\log_3(10)\log_{10}(12) = \log_3(12)\). 2. Apply the identity again: \(\log_3(12)\log_{12}(81) = \log_3(81)\). 3. Since \(81 = 3^4\), the value is \(4\).

\(4\)

- Express \(8\) and \(125\) as powers with the same exponent. - Simplify the first logarithm using those power forms. - Recall the relationship between \(\log_a(b)\) and \(\log_b(a)\).

1. Rewrite \(8 = 2^3\) and \(125 = 5^3\). 2. Then \(\log_8(125) = \log_{2^3}(5^3) = \log_2(5)\). 3. The product becomes \(\log_2(5)\log_5(2)\). 4. Since reciprocal logarithms multiply to \(1\), the value is \(1\).

\(1\)

- Use the change-of-base formula with common logarithms. - Organize the bases and computed values in increasing order. - Consider how the required exponent changes when the base grows but the resulting power remains \(20\).

1. By change of base, \(\log_2(20) = \frac{\log(20)}{\log(2)} \approx 4.32\). 2. \(\log_5(20) = \frac{\log(20)}{\log(5)} \approx 1.86\). 3. \(\log_{10}(20) = \log(20) \approx 1.30\). 4. Since \(4.32 > 1.86 > 1.30\) while \(2 < 5 < 10\), the computed values support the conjecture for the argument \(20\).

\(\log_2(20) \approx 4.32\), \(\log_5(20) \approx 1.86\), and \(\log_{10}(20) \approx 1.30\). These values decrease as the base increases.

- Rewrite roots as rational exponents. - Use the power property to move an exponent in front of a logarithm. - Use the quotient property to separate a numerator and denominator. - Express each number as a power related to the logarithm base.

1. a) Use the power property: \(\log_4(8^{1/2}) = \frac{1}{2}\log_4(8)\). Since \(4^{3/2} = 8\), the value is \(\frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}\). 2. b) \(\log_9(3^{1/3}) = \frac{1}{3}\log_9(3) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}\). 3. c) Use the quotient and power properties: \(\log_2(2^{1/3}) - \log_2(2^2) = \frac{1}{3} - 2 = -\frac{5}{3}\). 4. d) Use the quotient and power properties: \(\log_{10}(10^2) - \log_{10}(10^{1/4}) = 2 - \frac{1}{4} = \frac{7}{4}\).

a) \(\frac{3}{4}\) b) \(\frac{1}{6}\) c) \(-\frac{5}{3}\) d) \(\frac{7}{4}\)

- Move coefficients into exponents using the power property. - Combine added logarithms with the product property. - Distribute the negative sign before combining part b. - Combine a difference of logarithms with the quotient property.

1. a) Apply the power property: \(\log_a(x^3) + \log_a(y^2) - \log_a(z^4)\). 2. Use the product and quotient properties: \(\log_a\left(\frac{x^3y^2}{z^4}\right)\). 3. b) Combine the logarithms in parentheses: \(\log_k(v) + 2\log_k(w) = \log_k(vw^2)\). 4. Rewrite the first term and apply the quotient property: \(\log_k(u^5) - \log_k(vw^2) = \log_k\left(\frac{u^5}{vw^2}\right)\).

a) \(\log_a\left(\frac{x^3y^2}{z^4}\right)\) b) \(\log_k\left(\frac{u^5}{vw^2}\right)\)

- Move each coefficient into an exponent. - Rewrite the exponent \(\frac{1}{2}\) as a square root. - Condense each expression into one logarithm of a quotient. - Simplify common factors in the final argument.

1. a) Apply the power property: \(\frac{1}{2}\log_b(100p^2) = \log_b(10p)\). The original logarithms imply \(p > 0\). 2. Use the quotient property: \(\log_b(10p) - \log_b(2p) = \log_b\left(\frac{10p}{2p}\right) = \log_b(5)\). 3. b) Apply the power property: \(\log_x((4a)^2) + \log_x(b^3) - \log_x(8a^2b^3)\). 4. Combine and simplify: \(\log_x\left(\frac{16a^2b^3}{8a^2b^3}\right) = \log_x(2)\).

a) \(\log_b(5)\) b) \(\log_x(2)\)

- Begin with the outermost operation in each argument. - Rewrite roots as rational exponents. - Do not split a sum such as \(x+y\). - When expanding a reciprocal of a product, distribute the negative sign to every resulting logarithm.

1. a) Apply the product and power properties: \(\log_a(x^3y^5) = 3\log_a(x) + 5\log_a(y)\). 2. b) Apply the quotient and power properties: \(\log_a\left(\frac{b^2}{\sqrt{c}}\right) = 2\log_a(b) - \frac{1}{2}\log_a(c)\). 3. c) Rewrite the fourth root as a power of \(\frac{1}{4}\): \(\frac{1}{4}\left(3\log_a(u) - \log_a(v)\right) = \frac{3}{4}\log_a(u) - \frac{1}{4}\log_a(v)\). 4. d) Use the reciprocal, product, and power properties: \(-\log_a(z^2(x+y)) = -2\log_a(z) - \log_a(x+y)\).

a) \(3\log_a(x) + 5\log_a(y)\) b) \(2\log_a(b) - \frac{1}{2}\log_a(c)\) c) \(\frac{3}{4}\log_a(u) - \frac{1}{4}\log_a(v)\) d) \(-2\log_a(z) - \log_a(x+y)\)

- Move coefficients into exponents before combining logarithms. - Added logarithms produce a product, and subtracted logarithms produce a quotient. - Look for a difference-of-squares product. - In part d, condense the expression inside the parentheses before applying the outside coefficient.

1. a) Apply the power and product properties: \(\log_k(x^4) + \log_k(y^3) = \log_k(x^4y^3)\). 2. b) Apply the power and quotient properties: \(\log_k(\sqrt[3]{a}) - \log_k(b^2) = \log_k\left(\frac{\sqrt[3]{a}}{b^2}\right)\). 3. c) Use the product property, then the difference-of-squares identity: \(\log_k((u+v)(u-v)) = \log_k(u^2-v^2)\). 4. d) First condense inside the parentheses: \(\log_k\left(\frac{r}{s^2}\right)\). Then apply the outer coefficient: \(\log_k\left(\left(\frac{r}{s^2}\right)^3\right) = \log_k\left(\frac{r^3}{s^6}\right)\).

a) \(\log_k(x^4y^3)\) b) \(\log_k\left(\frac{\sqrt[3]{a}}{b^2}\right)\) c) \(\log_k(u^2-v^2)\) d) \(\log_k\left(\frac{r^3}{s^6}\right)\)

- Use the quotient property on the difference in part a. - Factor \(x^2-9\) as a difference of squares. - Expand the quotient logarithm in part b. - Combine like logarithm terms and simplify common factors.

1. a) Apply the quotient property: \(\log_a\left(\frac{x^2-9}{x+3}\right)\). 2. Factor the numerator and simplify: \(\log_a\left(\frac{(x-3)(x+3)}{x+3}\right) = \log_a(x-3)\). 3. b) Expand the last term: \(\log_a\left(\frac{q^3}{p}\right) = 3\log_a(q) - \log_a(p)\). 4. Combine like terms: \(2\log_a(p) - 3\log_a(q) + 3\log_a(q) - \log_a(p) = \log_a(p)\).

a) \(\log_a(x-3)\) b) \(\log_a(p)\)

- Rewrite the constant \(1\) as a logarithm with base \(5\). - Use the product property to combine the logarithms on the left. - Determine the domain before solving. - Equal logarithms with the same valid base have equal arguments.

1. The logarithm requires \(x-1>0\), so \(x>1\). 2. Rewrite \(1\) as \(\log_5(5)\): \(\log_5(x-1)+\log_5(5)=\log_5(20)\). 3. Apply the product property: \(\log_5(5(x-1))=\log_5(20)\). 4. Because logarithms with the same valid base are one-to-one, set the arguments equal: \(5(x-1)=20\). 5. Solve: \(x-1=4\), so \(x=5\). 6. Since \(5>1\), the solution is in the domain.

\(\{5\}\)

- Use the quotient property to combine the logarithms. - Rewrite the logarithmic equation in exponential form. - Determine the domain before solving. - Clear the fraction after converting to exponential form.

1. The logarithms require \(x+6>0\) and \(x>0\), so the domain is \(x>0\). 2. Apply the quotient property: \(\log_2\left(\frac{x+6}{x}\right)=2\). 3. Rewrite in exponential form: \(\frac{x+6}{x}=2^2=4\). 4. Solve: \(x+6=4x\), so \(6=3x\) and \(x=2\). 5. Since \(2>0\), the solution is in the domain.

\(\{2\}\)

- Combine logarithms that have the same base. - Rewrite a constant as a logarithm with the needed base. - Use the definition of a logarithm to convert between logarithmic and exponential form. - Check that every logarithm argument is positive.

1. a) Combine the logarithms: \(\log_3(4x)=\log_3(12)\). Set the arguments equal: \(4x=12\), so \(x=3\). Since \(3>0\), the solution is valid. 2. b) Rewrite \(1\) as \(\log(10)\): \(\log(x)=\log(2)+\log(10)=\log(20)\). Therefore, \(x=20\), which is in the domain. 3. c) Rewrite in exponential form: \(2x-4=5^2=25\). Then \(2x=29\), so \(x=14.5\). Its logarithm argument is \(25>0\), so the solution is valid.

a) \(\{3\}\) b) \(\{20\}\) c) \(\{14.5\}\)

- Determine the domain of the logarithm. - Use a logarithm property to combine the left side. - Rewrite \(2\) as a common logarithm or convert the inequality to exponential form. - Consider whether the base causes the inequality direction to stay the same or reverse.

1. The logarithm requires \(x>0\). 2. Apply the product property: \(\log(5x)\ge2\). 3. Because the common-logarithm base is \(10>1\), exponentiating preserves the inequality: \(5x\ge10^2\). 4. Thus, \(5x\ge100\), so \(x\ge20\). 5. These values satisfy the domain restriction, so the solution set is \([20, \infty)\).

\([20, \infty)\)

- Determine the domain first. - Rewrite \(1\) as a logarithm with base \(3\). - Use the quotient property to combine the logarithms. - Intersect the inequality result with the domain.

1. The logarithm requires \(x+4>0\), so \(x>-4\). 2. Rewrite \(1\) as \(\log_3(3)\): \(\log_3(x+4)-\log_3(3)<\log_3(2)\). 3. Apply the quotient property: \(\log_3\left(\frac{x+4}{3}\right)<\log_3(2)\). 4. Because the base \(3>1\), the logarithmic function is increasing, so \(\frac{x+4}{3}<2\). 5. Solve: \(x+4<6\), so \(x<2\). 6. Intersect this result with the domain to obtain \(-4<x<2\).

\((-4, 2)\)

- Determine which values of \(x\) are allowed in both logarithms. - Use the product property to combine logarithms with the same base. - Convert an equation of the form \(\log_b(A)=c\) to exponential form. - Check each solution of the quadratic equation against the original domain.

1. The logarithms require \(x+5>0\) and \(x>0\), so the domain is \(x>0\). 2. Apply the product property: \(\log_6(x(x+5))=2\). 3. Rewrite in exponential form: \(x(x+5)=6^2=36\). 4. Solve \(x^2+5x-36=0\): \((x-4)(x+9)=0\), so \(x=4\) or \(x=-9\). 5. Only \(x=4\) is in the domain.

\(\{4\}\)

- Determine when each logarithm argument is positive. - Convert a logarithmic equation to exponential form. - Use the product property when logarithms with the same base are added. - Check all calculated values in the domain of the original equation.

1. a) Convert to exponential form: \(4x-5=3^2=9\). Thus, \(4x=14\), so \(x=3.5\). The logarithm argument is \(9>0\), so the solution is valid. 2. b) Convert to exponential form: \(x^2+19=10^2=100\). Then \(x^2=81\), so \(x=\pm9\). Both values make the logarithm argument equal to \(100\), so both are valid. 3. c) The domain is \(x>2\). Apply the product property: \(\log_2(x(x-2))=3\). Therefore, \(x^2-2x=2^3=8\), so \((x-4)(x+2)=0\). Only \(x=4\) is in the domain.

a) \(\{3.5\}\) b) \(\{-9, 9\}\) c) \(\{4\}\)

- Determine when every logarithm argument is positive. - Use the quotient property to combine the logarithms on the left. - When logarithms with the same valid base are equal, compare their arguments. - Check the result in the original equation and domain.

1. The logarithms require \(x+4>0\) and \(x-1>0\), so the domain is \(x>1\). 2. Apply the quotient property: \(\log_5\left(\frac{x+4}{2}\right)=\log_5(x-1)\). 3. Equal logarithms with the same valid base have equal arguments, so \(\frac{x+4}{2}=x-1\). 4. Solve: \(x+4=2x-2\), so \(x=6\). 5. Since \(6>1\), the solution is in the domain.

\(\{6\}\)

- Relate common logarithms to powers of \(10\). - Rewrite a negative power of \(10\) as a decimal. - Which logarithm property applies when the argument is a product? - Replace \(x\) with \(10x\) and simplify.

1. a) \(h(10) = 1\) because \(10^1 = 10\). Similarly, \(h(100) = 2\) and \(h(1{,}000{,}000) = 6\). 2. b) Rewrite \(\log_{10}(x) = -4\) in exponential form: \(x = 10^{-4} = 0.0001\). 3. c) For \(x > 0\), the product property gives \(h(10x) = \log_{10}(10x) = \log_{10}(10) + \log_{10}(x) = 1 + h(x)\). Therefore, multiplying the input by \(10\) increases the output by \(1\).

a) \(h(10) = 1\), \(h(100) = 2\), and \(h(1{,}000{,}000) = 6\) b) \(x = 0.0001\) c) \(h(10x) = 1 + h(x)\) for \(x > 0\).

- First combine the logarithms inside parentheses. - Use the power property to move coefficients into exponents. - Rewrite fractional exponents as radicals. - A subtraction of logarithms produces a quotient.

1. a) Use the power and quotient properties: \(\log(x) = \frac{1}{3}\log\left(\frac{p^2}{q}\right)\). 2. Move the coefficient into the exponent: \(\log(x) = \log\left(\left(\frac{p^2}{q}\right)^{1/3}\right)\). 3. Therefore, \(x = \sqrt[3]{\frac{p^2}{q}}\). 4. b) First combine the logarithms in parentheses: \(\log(b) + 2\log(c) = \log(bc^2)\). 5. Apply the power property: \(\frac{3}{4}\log(bc^2) = \log\left((bc^2)^{3/4}\right) = \log\left(\sqrt[4]{b^3c^6}\right)\). 6. Use the quotient property: \(\log(x) = \log\left(\frac{a}{\sqrt[4]{b^3c^6}}\right)\). 7. Therefore, \(x = \frac{a}{\sqrt[4]{b^3c^6}}\).

a) \(x = \sqrt[3]{\frac{p^2}{q}}\) b) \(x = \frac{a}{\sqrt[4]{b^3c^6}}\)

- Express \(1\) as a common logarithm. - Simplify the logarithms inside the parentheses first. - Use the power property to rewrite a factor of \(\frac{1}{3}\) as a cube root. - Combine the final difference using the quotient property.

1. Combine the logarithms inside the parentheses: \(\log(u+v) + 2\log(u-v) = \log\left((u+v)(u-v)^2\right)\). 2. Apply the power property: \(\frac{1}{3}\log\left((u+v)(u-v)^2\right) = \log\left(\sqrt[3]{(u+v)(u-v)^2}\right)\). 3. Write \(1 = \log(10)\). 4. Apply the quotient property: \(\log(x) = \log\left(\frac{10}{\sqrt[3]{(u+v)(u-v)^2}}\right)\). 5. Therefore, \(x = \frac{10}{\sqrt[3]{(u+v)(u-v)^2}}\).

\(x = \frac{10}{\sqrt[3]{(u+v)(u-v)^2}}\)

- Use the power property for each coefficient of \(\frac{1}{2}\). - Combine the resulting logarithms with the product and quotient properties. - In part b, substitute \(3a\) for every occurrence of \(b\). - Use \(a > 0\) when simplifying \(\sqrt{a^2}\).

1. a) Apply the power property and combine the logarithms: \(\log(x) = \log(\sqrt{a}) + \log(\sqrt{a+b}) - \log(c)\). 2. Use the product and quotient properties: \(\log(x) = \log\left(\frac{\sqrt{a(a+b)}}{c}\right)\). 3. Therefore, \(x = \frac{\sqrt{a(a+b)}}{c}\). 4. b) Substitute \(b = 3a\): \(x = \frac{\sqrt{a(a+3a)}}{c} = \frac{\sqrt{4a^2}}{c}\). 5. Since \(a > 0\), \(\sqrt{4a^2} = 2a\). Thus, \(x = \frac{2a}{c}\).

a) \(x = \frac{\sqrt{a(a+b)}}{c}\) b) \(x = \frac{\sqrt{4a^2}}{c} = \frac{2a}{c}\)

- Distribute each coefficient carefully, including the negative sign before the second bracket. - Use the power property on each logarithm. - Positive logarithm terms become factors in the numerator, and negative terms become factors in the denominator. - Rewrite rational exponents as radicals at the end.

1. Distribute the coefficients: \(\log(x) = \frac{2}{3}\log(a+b) - \frac{1}{3}\log(a) - \frac{1}{4}\log(b) + \frac{1}{2}\log(a-b)\). 2. Apply the power property: \(\log(x) = \log\left((a+b)^{2/3}\right) - \log\left(a^{1/3}\right) - \log\left(b^{1/4}\right) + \log\left((a-b)^{1/2}\right)\). 3. Combine the logarithms: \(\log(x) = \log\left(\frac{(a+b)^{2/3}(a-b)^{1/2}}{a^{1/3}b^{1/4}}\right)\). 4. Therefore, \(x = \frac{\sqrt[3]{(a+b)^2}\sqrt{a-b}}{\sqrt[3]{a}\sqrt[4]{b}}\).

\(x = \frac{\sqrt[3]{(a+b)^2}\sqrt{a-b}}{\sqrt[3]{a}\sqrt[4]{b}}\)

- In part 1, combine the logarithms before simplifying the product. - Use the difference-of-squares identity. - In part 2, express \(1\) as a common logarithm. - Rewrite rational exponents as radicals.

1. For part 1, factor out \(\frac{1}{2}\) and use the product property: \(\log(x) = \frac{1}{2}\log\left((a+b)(a-b)\right)\). 2. Since \((a+b)(a-b) = a^2-b^2\), \(\log(x) = \log\left(\sqrt{a^2-b^2}\right)\). Therefore, \(x = \sqrt{a^2-b^2}\). 3. For part 2, write \(1 = \log(10)\) and combine the logarithms in parentheses: \(2\log(a) + \log(b) = \log(a^2b)\). 4. Apply the power property and then the quotient property: \(\log(x) = \log(10) - \log\left(\sqrt[3]{a^2b}\right) = \log\left(\frac{10}{\sqrt[3]{a^2b}}\right)\). 5. Therefore, \(x = \frac{10}{\sqrt[3]{a^2b}}\).

1) \(x = \sqrt{a^2-b^2}\) 2) \(x = \frac{10}{\sqrt[3]{a^2b}}\)

- Recall the definition of a composite number. - Use prime factorization to identify which logarithms are needed. - Derive \(\log(5)\) from \(\log(10)\) and \(\log(2)\). - Check each integer from \(1\) through \(20\) by its prime factors.

1. The claim is false. For example, \(14 = 2 \cdot 7\), so \(\log(14) = \log(2) + \log(7)\). The value of \(\log(7)\) is not provided and cannot be obtained from the stated values using logarithm properties alone. 2. Since \(\log(5) = \log(10) - \log(2)\), logarithms can be determined for numbers whose prime factorizations use only \(2\), \(3\), and \(5\). From \(1\) through \(20\), these are \(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18,\) and \(20\). 3. The missing primes are \(7, 11, 13, 17,\) and \(19\). Knowing their logarithms would allow every integer from \(1\) through \(20\) to be handled by prime factorization.

1. The claim is false; for example, finding \(\log(14)\) requires \(\log(7)\). 2. \(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20\) 3. \(7, 11, 13, 17, 19\)

- Use the power property to move a coefficient into a logarithm as an exponent. - Use the quotient property for a difference of logarithms. - Find the values excluded by the logarithm arguments. - A solution of a resulting quadratic equation may not satisfy the original domain.

1. a) The domain is \(x>3\). Combine the logarithms and convert to exponential form: \((x+3)(x-3)=2^4\). Thus, \(x^2-9=16\), so \(x=\pm5\). Only \(x=5\) is in the domain. 2. b) The domain is \(x>0\). Use the power property: \(\log(x^2)=\log(x+20)\). Therefore, \(x^2=x+20\), so \((x-5)(x+4)=0\). Only \(x=5\) is in the domain. 3. c) The domain is \(x>1\). Use the quotient property and convert to exponential form: \(\frac{x+2}{x-1}=4\). Then \(x+2=4x-4\), so \(x=2\), which is in the domain.

a) \(\{5\}\) b) \(\{5\}\) c) \(\{2\}\)

- Substitute the coordinates of \((1, 0)\) into the general equation. - Determine the exponent needed to raise each base to \(9\). - Use change of base to rewrite one function using the other function's base.

1. a) At \(x=1\), \(y=a\log_b(1)=a\cdot0=0\). Therefore, every graph passes through \((1, 0)\). 2. b) For \(a=2\) and \(b=3\), \(2\log_3(9)=2\cdot2=4\). 3. For \(a=4\) and \(b=9\), \(4\log_9(9)=4\cdot1=4\). Both parameter pairs produce a graph through \((9, 4)\). 4. c) Change the base of the second function to \(3\): \(4\log_9(x)=4\frac{\log_3(x)}{\log_3(9)}\). 5. Since \(\log_3(9)=2\), \(4\log_9(x)=2\log_3(x)\). The equations define the same function.

a) \(a\log_b(1)=0\), so every graph passes through \((1, 0)\). b) Both parameter pairs give an output of \(4\) when \(x=9\). c) They are two forms of the same function because \(4\log_9(x)=2\log_3(x)\).

- Use the change-of-base formula to express every logarithm with base \(2\). - Relate the bases \(\sqrt{2}\), \(2\), and \(4\). - Combine the logarithms so that each side has one logarithm with the same base. - Equal logarithms with the same valid base have equal arguments.

1. The domain requires \(x>0\). 2. Rewrite each logarithm with base \(2\): \(\log_4(9)=\frac{\log_2(9)}{\log_2(4)}=\frac{2\log_2(3)}{2}=\log_2(3)\). 3. Also, \(\log_{\sqrt{2}}(6)=\frac{\log_2(6)}{\log_2(\sqrt{2})}=\frac{\log_2(6)}{1/2}=2\log_2(6)=\log_2(36)\). 4. The equation becomes \(\log_2(x)+\log_2(3)=\log_2(36)\). 5. Apply the product property: \(\log_2(3x)=\log_2(36)\). Therefore, \(3x=36\), so \(x=12\). 6. Since \(12>0\), the solution is valid.

\(x=12\)