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: 68

Answer

$x=0;x=3$

Work Step by Step

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