Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.4 - Properties of Logarithms - 9.4 Exercises - Page 613: 21

Answer

$\dfrac{1}{2}\log_3 x +\dfrac{1}{2}\log_3 y-\dfrac{1}{2}\log_3 5$

Work Step by Step

Using the properties of logarithms, the given expression, $ \log_3\sqrt{\dfrac{xy}{5}} $, is equivalent to \begin{align*} & \log_3\left(\dfrac{xy}{5}\right)^{1/2} \\\\&= \dfrac{1}{2}\log_3 \dfrac{xy}{5} &(\text{use }\log_b x^y=y\log_b x) \\\\&= \dfrac{1}{2}\left(\log_3 (xy)-\log_3 5\right) &(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y) \\\\&= \dfrac{1}{2}\left(\log_3 x +\log_3 y-\log_3 5\right) &(\text{use }\log_b (xy)=\log_b x+\log_b y) \\\\&= \dfrac{1}{2}(\log_3 x) +\dfrac{1}{2}(\log_3 y)-\dfrac{1}{2}(\log_3 5) &(\text{use Distributive Property} \\\\&= \dfrac{1}{2}\log_3 x +\dfrac{1}{2}\log_3 y-\dfrac{1}{2}\log_3 5 .\end{align*} Hence, the expression $ \log_3\sqrt{\dfrac{xy}{5}} $ is equivalent to $ \dfrac{1}{2}\log_3 x +\dfrac{1}{2}\log_3 y-\dfrac{1}{2}\log_3 5 $.
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