College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3 - Page 477: 26

Answer

$2-\frac{1}{2}log_{8}(x+1)$

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, $log_{8}(\frac{64}{\sqrt (x+1)})=log_{8}(\frac{64}{(x+1)^{\frac{1}{2}}})=log_{8}64-log_{8}(x+1)^{\frac{1}{2}}$. 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, $log_{8}64-log_{8}(x+1)^{\frac{1}{2}}=log_{8}64-\frac{1}{2}log_{8}(x+1)$. 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, $log_{8}64=2$, because $8^{2}=64$. So, $log_{8}64-\frac{1}{2}log_{8}(x+1)=2-\frac{1}{2}log_{8}(x+1)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.