Answer
$3$
Work Step by Step
$\log _{16}\sqrt {x+1}=\dfrac {1}{4}=\sqrt {x+1}=16^{\dfrac {1}{4}}=2\Rightarrow x+1=2^{2}\Rightarrow x=2^{2}-1=3$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
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
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