Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.5 - Logarithmic Functions - Exercise Set - Page 875: 36

Answer

2

Work Step by Step

We know that if $b\gt0$ and $b\ne1$, then $y=log_{b}x$ is equivalent to $x=b^{y}$ (where $x\gt0$ and $y$ is a real number). Therefore, $log_{\frac{2}{3}}\frac{4}{9}=2$, because $(\frac{2}{3})^{2}=\frac{2^{2}}{3^{2}}=\frac{4}{9}$.
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