Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.4 - Properties of Logarithms - 9.4 Exercises: 20

Answer

$\frac{1}{3}log_{3}13-log_{3}p-2log_{3}q$

Work Step by Step

We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$). Therefore, $log_{3}\frac{\sqrt[3] 13}{pq^{2}}=log_{3}\sqrt[3] 13-log_{3}pq^{2}$. We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$). Therefore, $log_{3}\sqrt[3] 13-log_{3}pq^{2}=log_{3}\sqrt[3] 13-(log_{3}p+log_{3}q^{2})=log_{3}\sqrt[3] 13-log_{3}p-log_{3}q^{2}$. We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number). Therefore, $log_{3}\sqrt[3] 13-log_{3}p-log_{3}q^{2}=log_{3}13^{\frac{1}{3}}-log_{3}p-log_{3}q^{2}=\frac{1}{3}log_{3}13-log_{3}p-2log_{3}q$.
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.