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 - Page 98: 10

Answer

$2^{-3}=\displaystyle \frac{1}{8}$

Work Step by Step

From the definition of logarithms, for $a>0$, $a\neq 1$, and $x > 0$, $y=\log_{a}x$ means $a^{y}=x$. --------- $-3=\displaystyle \log_{2}(\frac{1}{8})\ \ $ means $\ \ 2^{-3}=\displaystyle \frac{1}{8} $.
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