Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises - Page 98: 29

Answer

$1+\log_{3}p-\log_{3}5-\log_{3}k$

Work Step by Step

...Apply rule: $\displaystyle \log_{a}\frac{x}{y}=\log_{a}x-\log_{a}y$ $=\displaystyle \log_{3}(\frac{3p}{5k})$ $=\log_{3}(3\cdot p)-\log_{3}(5\cdot k)$ ...Apply rule: $\log_{a}xy=\log_{a}x+\log_{a}y$ $=\log_{3}3+\log_{3}p-(\log_{3}5+\log_{3}k)$ ... $\log_{b}b=1$, remove parentheses. $=1+\log_{3}p-\log_{3}5-\log_{3}k$
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