College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.3: 68

Answer

$=\ln \sqrt[3] {(\frac{(x+6)^{5}}{x(x^{2}-25)})}$

Work Step by Step

$= \frac{1}{3}[5\ln(x+6)-\ln(x)-\ln(x^{2}-25)]$ $= \frac{1}{3}[\ln(x+6)^{5}-\ln x - \ln (x^{2}-25)]$ $= \frac{1}{3}[\ln(\frac{(x+6)^{5}}{x})- \ln (x^{2}-25)]]$ $= \frac{1}{3}[\ln(\frac{(x+6)^{5}}{x}\times\frac{1}{x^{2}-25})]$ $=\frac{1}{3}[\ln(\frac{(x+6)^{5}}{x(x^{2}-25)})]$ $= \ln(\frac{(x+6)^{5}}{x(x^{2}-25)})^{\frac{1}{3}}$ $=\ln \sqrt[3] {(\frac{(x+6)^{5}}{x(x^{2}-25)})}$
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