## College Algebra (6th Edition)

Published by Pearson

# Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 477: 23

#### Answer

$\frac{1}{2}log_{4}x-3$

#### Work Step by Step

Based on the quotient rule of logarithms, we know that $log_{b}(\frac{M}{N})=log_{b}M-log_{b}N$ (where $b$, $M$, and $N$ are positive real numbers and $b\ne1$). Therefore, $log_{4}(\frac{\sqrt x}{64})=log_{4}(\frac{x^{\frac{1}{2}}}{64})=log_{4}x^{\frac{1}{2}}-log_{4}64$. According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number). Therefore, $log_{4}x^{\frac{1}{2}}-log_{4}64=\frac{1}{2}log_{4}x-log_{4}64$. Based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$). Therefore, $log_{4}64=3$, because $4^{3}=64$. So, $\frac{1}{2}log_{4}x-log_{4}64=\frac{1}{2}log_{4}x-3$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.