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: 65

Answer

$x_{1}=0;x_{2}=-6$

Work Step by Step

$\log _{8}\left( x+2\right) +\log _{8}\left( x+4\right) =\log _{8}8\Rightarrow \log _{8}\left( \left( x+2\right) \left( x+4\right) \right) =\log _{8}8$ $\Rightarrow x^{2}+6x+8=8\Rightarrow x^{2}+6x=0\Rightarrow x_{1}=0;x_{2}=-6$
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