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: 75

Answer

$x=\pm 3$

Work Step by Step

$\log _{5}\left( x+2\right) +\log _{5}\left( x-2\right) =1\Rightarrow \log _{5}\left( \left( x+2\right) \left( x-2\right) \right) =\log _{5}5$ $\Rightarrow \left( x+2\right) \left( x-2\right) =5\Rightarrow x^{2}-4=5\Rightarrow x^{2}=9\Rightarrow x=\pm 3$
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