Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.5 Exponential and Logarithmic Equations - 4.5 Exercises - Page 469: 76

Answer

$x_{1}=8;x_{2}=-1$

Work Step by Step

$\log _{2}\left( x-7\right) +\log _{2}x=3\Rightarrow \log _{2}\left( \left( x-7\right) x\right) =\log _{2}8$ $\Rightarrow x\left( x-7\right) =8\Rightarrow x^{2}-7x-8=0\Rightarrow \left( x-8\right) \left( x+1\right) =0\Rightarrow x_{1}=8;x_{2}=-1$
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