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: 21

Answer

$\frac{7}{2}$

Work Step by Step

Use quotient rule, we have $log_3(\frac{x+1}{x-3})=2\longrightarrow \frac{x+1}{x-3}=3^2=9\longrightarrow x+1=9x-27\longrightarrow 8x=28$, thus $x=\frac{7}{2}$
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