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 614: 36

Answer

$\log_b \dfrac{x^{1/3}y^{2/3}}{s^{3/4}t^{2/3}}$

Work Step by Step

Using the properties of logarithms, the given expression, $ \dfrac{1}{3}\log_b x+\dfrac{2}{3}\log_b y-\dfrac{3}{4}\log_b s-\dfrac{2}{3}\log_b t $, is equivalent to \begin{align*} & \log_b x^{\frac{1}{3}}+\log_b y^{\frac{2}{3}}-\log_b s^{\frac{3}{4}}-\log_b t^{\frac{2}{3}} &(\text{use }\log_b x^y=y\log_b x) \\\\&= \left(\log_b x^{\frac{1}{3}}+\log_b y^{\frac{2}{3}}\right)-\left(\log_b s^{\frac{3}{4}}-\log_b t^{\frac{2}{3}}\right) \\\\&= \log_b \left(x^{\frac{1}{3}}y^{\frac{2}{3}}\right)-\log_b \left(s^{\frac{3}{4}}t^{\frac{2}{3}}\right) &(\text{use }\log_b (xy)=\log_b x+\log_b y) \\\\&= \log_b \dfrac{x^{\frac{1}{3}}y^{\frac{2}{3}}}{s^{\frac{3}{4}}t^{\frac{2}{3}}} &(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y) \\\\&= \log_b \dfrac{x^{1/3}y^{2/3}}{s^{3/4}t^{2/3}} .\end{align*} Hence, the expression $ \dfrac{1}{3}\log_b x+\dfrac{2}{3}\log_b y-\dfrac{3}{4}\log_b s-\dfrac{2}{3}\log_b t $ is equivalent to $ \log_b \dfrac{x^{1/3}y^{2/3}}{s^{3/4}t^{2/3}} $.
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