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: 53


$\log ( \frac{x^4(x^2-1)}{\sqrt [3] {x^2+1}})$

Work Step by Step

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