Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises: 47


$\frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$

Work Step by Step

$Expand$ $the$ $logarithmic$ $expression:$ $\ln \sqrt {\frac{x^2-1}{x^2+1}}$ Rewrite the root to exponent form $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}}$$ Use the Third Law of Logarithms $$\ln (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} = \frac{1}{2}\ln (\frac{x^2-1}{x^2+1})$$ Use the Second Law of Logarithms [Distribute the $\frac{1}{2}$ as well] $$\frac{1}{2}\ln (\frac{x^2-1}{x^2+1}) = \frac{1}{2}\ln (x^2-1) - \frac{1}{2}\ln (x^2+1)$$
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