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: 29

Answer

$2$

Work Step by Step

Step 1. Rewrite the equation as $log_{1/3}(x^2+x)=log_{1/3}(x^2-x)-1 \longrightarrow log_{1/3}(x^2+x)=log_{1/3}(x^2-x)-log_{1/3}(1/3)$ Step 2. Thus $x^2+x=3(x^2-x)\longrightarrow 2x^2-4x=0\longrightarrow x=0,2$. Step 3. Check answers, only $x=2$ fits the original equation.
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