Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises - Page 359: 37


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

Work Step by Step

$Expand$ $the$ $expression$: $\log_3 \frac{\sqrt{3x^5}}{y}$ Apply the Second Law of Logarithms $\log_3 \frac{\sqrt{3x^5}}{y}$ = $\log_3 \sqrt{3x^5}$ - $\log_3 y$ Rewrite the square root for 3x$^5$ $\log_3 (3x^5)^\frac{1}{2}$ - $\log_3 y$ Apply the Third Law of Logarithms for $\log_3 (3x^5)^\frac{1}{2}$ $\log_3 (3x^5)^\frac{1}{2}$ = $\frac{1}{2}$$\log_3 3x^5$ Apply the First Law of Logarithms for $\frac{1}{2}$$\log_3 3x^5$ (Also distribute the half as well) $\frac{1}{2}$$\log_3 (3\times x^5)$ = $\frac{1}{2}$$\log_3 3$ + $\frac{1}{2}$$\log_3 x^5$ Use the Property $\log_x x$ $=$ $1$ for $\frac{1}{2}$$\log_3 3$ $\frac{1}{2}$$\log_3 3$ = $\frac{1}{2}$ $\times$ $1$ = $\frac{1}{2}$ Apply the Third Law of Logarithms for $\frac{1}{2}$$\log_3 x^5$ $\frac{1}{2}$$\log_3 x^5$ = $\frac{5}{2}$$\log_3 x$ [Note: $\frac{1}{2}$ $\times$ 5 = $\frac{5}{2}$] Assemble the expression back together (Don't forget the $\log_3 y$) $\frac{1}{2}$ + $\frac{5}{2}$$\log_3 x$ - $\log_3 y$
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