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: 19

Answer

$-\displaystyle \frac{2}{3}$

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_{2}(\sqrt[3]{\frac{1}{4}})\ \ $ means $\ \ 2^{y}=\displaystyle \sqrt[3]{\frac{1}{4}} .$ Since $\displaystyle \sqrt[3]{\frac{1}{4}}=(\frac{1}{4})^{1/3}=(4^{-1})^{1/3}$ $=(2^{2})^{-1/3}=2^{-2/3},$ $y=-\displaystyle \frac{2}{3}$
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