## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 3 - Review Exercises - Page 513: 52

#### Answer

$\log_{2}x+2\log_{2}y-6$

#### Work Step by Step

Basic logarithmic properties:$\left\{\begin{array}{l} \log_{b}1=0\\ \log_{b}b=1\\ \log_{b}b^{x}=x\\ b^{\log_{b}}x=x \end{array}\right.$ Rules: The Product Rule: $\log_{b}(MN)=\log_{b}\mathrm{M}+\log_{b}\mathrm{N}$ The Quotient Rule: $\displaystyle \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$ The Power Rule: $\log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$ --- $\displaystyle \log_{2}\frac{xy^{2}}{64}=\qquad$... apply Quotient Rule $=\log_{2}(xy^{2})-\log_{2}64=\qquad$... apply Product Rule $=\log_{2}x+\log_{2}y^{2}-\log_{2}64\qquad$... recognize $64=2^{6}$ $=\log_{2}x+\log_{2}y^{2}-\log_{2}2^{6}\qquad$... apply Power Rule $=\log_{2}x+2\log_{2}y-6\log_{2}2\qquad$... apply $\log_{b}b=1$ $=\log_{2}x+2\log_{2}y-6\cdot 1$ = $\log_{2}x+2\log_{2}y-6$

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