College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3: 4

Answer

$1+log_{9}x$

Work Step by Step

Based on the product rule of logarithms, we know that $log_{b}(MN)=log_{b}M+log_{b}N$ (for $M\gt0$ and $N\gt0$). Therefore, $log_{9}(9x)=log_{9}9+log_{9}x$. Based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$). Therefore, $log_{9}9=1$, because $9^{1}=9$. So, $log_{9}9+log_{9}x=1+log_{9}x$.
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