## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 12 - Exponential Functions and Logarithmic Functions - 12.4 Properties of Logarithmic Functions - 12.4 Exercise Set - Page 810: 43

#### Answer

$\dfrac{7}{2}\log_a x-\dfrac{5}{2}\log_a y-4\log_az$

#### Work Step by Step

Using the properties of logarithms, the given expression, $\log_a \sqrt{\dfrac{x^7}{y^5z^8}} ,$ is equivalent to \begin{array}{l}\require{cancel} \log_a \left( \dfrac{x^7}{y^5z^8} \right)^{1/2} \\\\= \dfrac{1}{2}\log_a \dfrac{x^7}{y^5z^8} \\\\= \dfrac{1}{2}\left[ \log_a x^7-\log_a (y^5z^8) \right] \\\\= \dfrac{1}{2}\left[ \log_a x^7-\left( \log_a y^5+\log_az^8 \right) \right] \\\\= \dfrac{1}{2}\left[ \log_a x^7-\log_a y^5-\log_az^8 \right] \\\\= \dfrac{1}{2}\left[ 7\log_a x-5\log_a y-8\log_az \right] \\\\= \dfrac{1}{2}\cdot7\log_a x-\dfrac{1}{2}\cdot5\log_a y-\dfrac{1}{2}\cdot8\log_az \\\\= \dfrac{7}{2}\log_a x-\dfrac{5}{2}\log_a y-\dfrac{8}{2}\log_az \\\\= \dfrac{7}{2}\log_a x-\dfrac{5}{2}\log_a y-4\log_az .\end{array}

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