## College Algebra (6th Edition)

Published by Pearson

# Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 477: 24

#### Answer

$\frac{1}{2}log_{5}x-2$

#### 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, $log_{5}(\frac{\sqrt x}{25})=log_{5}(\frac{x^{\frac{1}{2}}}{25})=log_{5}x^{\frac{1}{2}}-log_{5}25$. 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, $log_{5}x^{\frac{1}{2}}-log_{5}25=\frac{1}{2}log_{5}x-log_{5}25$. 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, $log_{5}25=2$, because $5^{2}=25$. So, $\frac{1}{2}log_{5}x-log_{5}25=\frac{1}{2}log_{5}x-2$.

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