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 - Page 466: 28

Answer

-2

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, $log_{3}\frac{1}{9}=-2$, because $3^{-2}=\frac{1}{3^{2}}=\frac{1}{3\times3}=\frac{1}{9}$.
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