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: 67

Answer

$12$

Work Step by Step

$\log _{2}\left( x^{2}-100\right) -\log _{2}\left( x+10\right) =1\Rightarrow \log _{2}\left( \dfrac {x^{2}-10^{2}}{x+10}\right) =1$ $\dfrac {\left( x-10\right) \left( x+10\right) }{\left( x+10\right) }=2^{1}\Rightarrow x-10=2\Rightarrow x=12$
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