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: 62



Work Step by Step

$\log _{7}\left( 3x+2\right) -\log _{7}\left( x-2\right) =1\Rightarrow \log _{7}\dfrac {\left( 3x+2\right) }{\left( x-2\right) }=1\Rightarrow \dfrac {3x+2}{\left( x-2\right) }=7^{1}$ $\Rightarrow 3x+2=7\left( x-2\right) \Rightarrow 3x+2=7x-14\Rightarrow 4x=16\Rightarrow x=4$
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