Answer
$\ln\sqrt {x^2(x+2)}=\ln x+\frac{1}{2}\ln(x+2)$
Work Step by Step
$\ln\sqrt {x^2(x+2)}=\ln{[x^2(x+2)}]^{\frac{1}{2}}=\frac{1}{2}\ln{x^2(x+2)}=\frac{1}{2}[\ln x^2+\ln(x+2)]=\frac{1}{2}[2\ln x+\ln(x+2)]=\ln x+\frac{1}{2}\ln(x+2)$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 5 - 5.3 - Properties of Logarithms - 5.3 Exercises - Page 385: 60
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