## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1105: 97

#### Answer

The required solution is ${{\log }_{7}}\left( \frac{\sqrt[5]{x}}{49{{y}^{10}}} \right)=\frac{1}{5}{{\log }_{7}}x-2-10{{\log }_{7}}y$.

#### Work Step by Step

\begin{align} & {{\log }_{7}}\left( \frac{{{x}^{\frac{1}{5}}}}{49{{y}^{10}}} \right)={{\log }_{7}}{{x}^{\frac{1}{5}}}-{{\log }_{7}}\left( 49{{y}^{10}} \right) \\ & ={{\log }_{7}}{{x}^{\frac{1}{5}}}-\left( {{\log }_{7}}49+10{{\log }_{7}}y \right) \\ & =\frac{1}{5}{{\log }_{7}}x-{{\log }_{7}}49-10{{\log }_{7}}y \end{align} ${{\log }_{7}}49=2$ because 2 is the power to which 7 must be raised to get 49. Thus, ${{\log }_{7}}\left( \frac{\sqrt[5]{x}}{49{{y}^{10}}} \right)=\frac{1}{5}{{\log }_{7}}x-2-10{{\log }_{7}}y$.

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