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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - Review Exercises: Chapter 12 - Page 843: 34

Answer

The equivalent expression for $\frac{1}{3}\left[ {{\log }_{a}}x-2{{\log }_{a}}y \right]$ is ${{\log }_{a}}\sqrt[3]{\left( \frac{x}{{{y}^{2}}} \right)}$.

Work Step by Step

$\frac{1}{3}\left[ {{\log }_{a}}x-2{{\log }_{a}}y \right]$ Apply the power rule for logarithms as follows. $\frac{1}{3}\left[ {{\log }_{a}}x-2{{\log }_{a}}y \right]=\frac{1}{3}\left[ {{\log }_{a}}x-{{\log }_{a}}{{y}^{2}} \right]$ Apply the quotient rule for logarithms as follows. $\frac{1}{3}\left[ {{\log }_{a}}x-2{{\log }_{a}}y \right]=\frac{1}{3}\left( {{\log }_{a}}\frac{x}{{{y}^{2}}} \right)$ Apply the power rule for logarithms as follows. $\begin{align} & \frac{1}{3}\left[ {{\log }_{a}}x-2{{\log }_{a}}y \right]={{\log }_{a}}{{\left( \frac{x}{{{y}^{2}}} \right)}^{\frac{1}{3}}} \\ & ={{\log }_{a}}\sqrt[3]{\left( \frac{x}{{{y}^{2}}} \right)} \end{align}$
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