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: 71


$-\dfrac {1995}{991}$

Work Step by Step

$\log \left( 9x+5\right) =3+\log \left( x+2\right) \Rightarrow \log _{10}\left( 9x+5\right) =\log _{10}1000+\log \left( x+2\right) $ $\Rightarrow \log _{10}\left( 9x+5\right) =\log _{10}\left( 1000\left( x+2\right) \right) \Rightarrow 9x+5=1000x+2000\Rightarrow x=-\dfrac {1995}{991}$
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