Calculus with Applications (10th Edition)

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

Chapter 2 - Nonlinear Functions - Chapter Review - Review Exercises - Page 114: 64

Answer

$0.8$

Work Step by Step

For $a>0, a\neq 1$, and $x>0,\qquad y=\log_{a}x$ means $a^{y}=x$. $.................................$ $\log_{32}16=x$ means $32^{x}=16$, $32^{x}=16$ $(2^{5})^{x}=16$ $2^{5x}=2^{4}\qquad \color{blue}{\text{ ...equate the exponents } }$ $5x=4$ $x=\displaystyle \frac{4}{5}=0.8$ With change of base, $\displaystyle \log_{a}x=\frac{\log_{b}x}{\log_{b}a}=\frac{\ln x}{\ln a}$, using a calculator, $\log_{32}16$=$\displaystyle \frac{\ln 16}{\ln 32}=0.8$
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