## College Algebra (6th Edition)

Published by Pearson

# Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 477: 25

#### Answer

$2-\frac{1}{2}log_{6}(x+1)$

#### 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, $log_{6}(\frac{36}{\sqrt (x+1)})=log_{6}(\frac{36}{(x+1)^{\frac{1}{2}}})=log_{6}36-log_{6}(x+1)^{\frac{1}{2}}$. 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, $log_{6}36-log_{6}(x+1)^{\frac{1}{2}}=log_{6}36-\frac{1}{2}log_{6}(x+1)$. 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, $log_{6}36=2$, because $6^{2}=36$. So, $log_{6}36-\frac{1}{2}log_{6}(x+1)=2-\frac{1}{2}log_{6}(x+1)$.

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