Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises: 20

Answer

$-\displaystyle \frac{1}{12}$

Work Step by Step

From the definition of logarithms, for $a>0$, $a\neq 1$, and $x > 0$, $y=\log_{a}x$ means $a^{y}=x$. --------- $y=\displaystyle \log_{8}(\sqrt[4]{\frac{1}{2}})\ \ $ means $\ \ 8^{y}=\displaystyle \sqrt[4]{\frac{1}{2}} .$ Since $\displaystyle \sqrt[4]{\frac{1}{2}}=(\frac{1}{2})^{1/4}=(2^{-1})^{1/4}$ ... now, since $2^{3}=8$, it follows that $2=8^{1/3},$ $=(8^{1/3})^{-1/4}=8^{-1/12},$ $y=-\displaystyle \frac{1}{12}$
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.