College Algebra (11th Edition)

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

Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 438: 93

Answer

$\dfrac{1}{2}(3u-5v)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to express the given expression, $ \ln \sqrt{\dfrac{a^3}{b^5}} ,$ in a form that uses $\ln a$ and $\ln b$ only. Then substitute $u$ for any instance of $\ln a$ and substitute $v$ for any instance of $\ln b.$ $\bf{\text{Solution Details:}}$ 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} \ln \left(\dfrac{a^3}{b^5}\right)^{1/2} .\end{array} 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} \dfrac{1}{2}\ln \left(\dfrac{a^3}{b^5}\right) .\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} \dfrac{1}{2}(\ln a^3-\ln b^5) .\end{array} 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} \dfrac{1}{2}(3\ln a-5\ln b) .\end{array} Substituting $u$ with $\ln a$ and $v$ with $\ln b,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2}(3u-5v) .\end{array}
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