Intermediate Algebra (12th Edition)

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

Chapter 9 - Section 9.3 - Logarithmic Functions - 9.3 Exercises: 13

Answer

$log_{8}\frac{1}{4}=-\frac{2}{3}$

Work Step by Step

For all positive numbers $a$, where $a\ne1$, and all positive numbers $x$, we know that $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $8^{-\frac{2}{3}}=\frac{1}{4}$ means the same as $log_{8}\frac{1}{4}=-\frac{2}{3}$.
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.