Precalculus (6th Edition)

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

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Test - Page 496: 17

Answer

$\frac{3}{4}$

Work Step by Step

By definition, from $log_x(\frac{9}{16})=2$, ($x\gt0, x\ne1$), we have $x^2=\frac{9}{16}$, thus $x=\frac{3}{4}$
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