Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises: 54

Answer

$$\log_5 (\frac{2(x+1)}{(3x+7)^{\frac{1}{3}}})$$ Simplified: $$\log_5 (\frac{2x+2}{(3x+7)^{\frac{1}{3}}})$$

Work Step by Step

$Combine$ $into$ $a$ $single$ $logarithm:$ $\log_5 2$ +$\log_5 (x+1)$ - $\frac{1}{3}\log_5 (3x+7)$ Use the First Law of Logarithms for $\log_5 2$ +$\log_5 (x+1)$ $\log_5 2$ +$\log_5 (x+1)$ = $\log_5 (2\times (x+1))$ $\log_5 2(x+1)$ - $\frac{1}{3}\log_5 (3x+7)$ Use the Third Law of Logarithms for $\frac{1}{3}\log_5 (3x+7)$ $\frac{1}{3}\log_5 (3x+7)$ = $\log_5 (3x+7)^{\frac{1}{3}}$ $\log_5 2(x+1)$ - $\log_5 (3x+7)^{\frac{1}{3}}$ Use the Second Law of Logarithms $\log_5 2(x+1)$ - $\frac{1}{3}\log_5 (3x+7)$ = $\log_5 \frac{2(x+1)}{(3x+7)^{\frac{1}{3}}}$ $$\log_5 (\frac{2(x+1)}{(3x+7)^{\frac{1}{3}}})$$ Simplify if necessary $$\log_5 (\frac{2x+2}{(3x+7)^{\frac{1}{3}}})$$
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