Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.6 Logarithmic and Exponential Equations - 5.6 Assess Your Understanding - Page 311: 16

Answer

$5$.

Work Step by Step

Step 1. Rewrite the equation as $log_3(x+4)=\frac{1}{2}(2+log_33^2)=\frac{1}{2}(2+2)=2$ Step 2. Thus $x+4=3^2=9$ and $x=5$.
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