Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.3 Logarithmic Functions - 4.3 Exercises - Page 445: 27

Answer

$9$

Work Step by Step

$x=2^{\log _{2}9}\Rightarrow \log _{2}x=\log _{2}2^{\log _{2}9}\Rightarrow \log _{2}x=\log _{2}9\Rightarrow 2^{x}=2^{9}\Rightarrow x=9$
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