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: 55

Answer

$\ln (\frac{a^2-b^2}{c^2}$)

Work Step by Step

$Combine$ $the$ $expression$: $\ln (a+b)$ + $\ln (a-b)$ - $2$$\ln c$ Apply the Third Law of Logarithms for $2$$\ln c$ $2$$\ln c$ = $\ln c^2$ $\ln (a+b)$ + $\ln (a-b)$ - $\ln c^2$ Apply the First Law of Logarithms for $\ln (a+b)$ + $\ln (a-b)$ $\ln (a+b)$ + $\ln (a-b)$ = $\ln ((a+b)(a-b))$ [Note: Use FOIL Method] $\ln (a^2-ab+ab-b^2)$ = $\ln (a^2-b^2)$ $\ln (a^2-b^2)$ - $\ln c^2$ Apply the Second Law of Logarithms $\ln (a^2-b^2)$ - $\ln c^2$ = $\ln (\frac{a^2-b^2}{c^2}$)
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