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: 52

Answer

$\ln (\frac{8x^2}{\sqrt{x+4}})$

Work Step by Step

$Combine$ $the$ $expression$: $3$$\ln 2$ + $2$$\ln x$ - $\frac{1}{2}$$\ln (x+4)$ Apply the Third Law of Logarithms for $3$$\ln 2$, $2$$\ln x$, and $\frac{1}{2}$$\ln (x+4)$ $3$$\ln 2$ = $\ln 2^3$ = $\ln 8$ $2$$\ln x$ = $\ln x^2$ $\frac{1}{2}$$\ln (x+4)$ = $\ln (x+4)^\frac{1}{2}$ $\ln 8$ + $\ln x^2$ - $\ln (x+4)^\frac{1}{2}$ $\ln 8$ + $\ln x^2$ - $\ln \sqrt{x+4}$ Apply the First Law of Logarithms for $\ln 8$ + $\ln x^2$ $\ln 8$ + $\ln x^2$ = $\ln (8\times x^2)$ $\ln 8x^2$ - $\ln \sqrt{x+4}$ Apply the Second Law of Logarithms $\ln 8x^2$ - $\ln \sqrt{x+4}$ = $\ln (\frac{8x^2}{\sqrt{x+4}})$
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