College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises - Page 424: 83

Answer

$\log_5 \dfrac{\sqrt[3]{5m^2}}{m}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to write the given expression, $ -\dfrac{2}{3}\log_5 5m^2+\dfrac{1}{2}\log_5 25m^2 ,$ as a single logarithm. $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} -\log_5 (5m^2)^{2/3}+\log_5 (25m^2)^{1/2} \\\\= \log_5 (25m^2)^{1/2}-\log_5 (5m^2)^{2/3} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_5 \dfrac{(25m^2)^{1/2}}{(5m^2)^{2/3}} \\\\= \log_5 \dfrac{(5^2m^2)^{1/2}}{(5m^2)^{2/3}} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_5 \dfrac{5^{2\cdot\frac{1}{2}}m^{2\cdot\frac{1}{2}}}{5^{\frac{2}{3}}m^{2\cdot\frac{2}{3}}} \\\\= \log_5 \dfrac{5^{1}m^{1}}{5^{\frac{2}{3}}m^{\frac{4}{3}}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \log_5 5^{1-\frac{2}{3}}m^{1-\frac{4}{3}} \\\\= \log_5 5^{\frac{3}{3}-\frac{2}{3}}m^{\frac{3}{3}-\frac{4}{3}} \\\\= \log_5 5^{\frac{1}{3}}m^{-\frac{1}{3}} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_5 \dfrac{5^{\frac{1}{3}}}{m^{\frac{1}{3}}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_5 \dfrac{\sqrt[3]{5^1}}{\sqrt[3]{m^1}} \\\\= \log_5 \dfrac{\sqrt[3]{5}}{\sqrt[3]{m}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} \log_5 \dfrac{\sqrt[3]{5^1}}{\sqrt[3]{m^1}} \\\\= \log_5 \dfrac{\sqrt[3]{5}}{\sqrt[3]{m}}\cdot\dfrac{\sqrt[3]{m^2}}{\sqrt[3]{m^2}} \\\\= \log_5 \dfrac{\sqrt[3]{5m^2}}{\sqrt[3]{m^3}} \\\\= \log_5 \dfrac{\sqrt[3]{5m^2}}{m} .\end{array}
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