Answer
See below
Work Step by Step
Let $x=\log_b m\\y=\log_b n$
Rewrite as: $b^x=m\\b^y=n$
Obtain: $\log_b \frac{b^x}{b^y}=\log_b b^{x-y}=(x-y)\log_b b$
Thus, $x-y=\log_b m+\log_b n\\
\rightarrow \log_b \frac{m}{n}=\log_b m-\log_b n$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 7 Exponential and Logarithmic Functions - 7.5 Apply Properties of Logarithms - 7.5 Exercises - Skill Practice - Page 511: 66
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