Precalculus (6th Edition)

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: 61


$x_{1}=\dfrac {10}{3};x_{2}=1$

Work Step by Step

$\log x+\log \left( 13-3x\right) =1\Rightarrow \log \left[ x\left( 13-3x\right) \right] =1$ $\Rightarrow x\left( 13-3x\right) =10^{1}\Rightarrow 3x^{2}-13x+10=0$ $\left( 3x-10\right) \left( x-1\right) =0\Rightarrow x_{1}=\dfrac {10}{3};x_{2}=1$
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