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 4.2: 102

Answer

$x=21$

Work Step by Step

From the definition of the logarithmic function on page 456, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$, $b\gt0$, and $b\ne1$). Therefore, $log_{5}(x+4)=2$ is equivalent to $5^{2}=(x+4)$ in logarithmic form. We know that $5^{2}=25$, so we can set $x+4=25$. Subtract 4 from both sides. $x=21$
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