Precalculus (6th Edition)

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

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Chapter 4 Test Prep - Review Exercises - Page 492: 64



Work Step by Step

$\log _{16}\sqrt {x+1}=\dfrac {1}{4}=\sqrt {x+1}=16^{\dfrac {1}{4}}=2\Rightarrow x+1=2^{2}\Rightarrow x=2^{2}-1=3$
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