College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check: 47

Answer

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

Work Step by Step

Use the rule $\sqrt{a} = a^{\frac{1}{2}}$ to obtain: $=\ln{\left(\dfrac{x^2-1}{x^2+1}\right)^{\frac{1}{2}}}$ RECALL: (1) $\ln{(P^n)}=n \cdot \ln{P}$ (2) $\ln{(PQ)}=\ln{P} + \ln{Q}$ (3) $\ln{(\frac{P}{Q})} = \ln{P} - \ln{Q}$ Use rule (1) above to obtain $=\frac{1}{2}\ln{\left(\dfrac{x^2-1}{x^2+1}\right)}$ Use rule (3) above to obtain: $=\frac{1}{2}[\ln{(x^2-1)-\ln{(x^2+1)}}]$ Since $a^2-b^2=(a-b)(a+b)$, then the expression above is equivalent to: $=\frac{1}{2}[\ln{[(x-1)(x+1)]}-\ln{(x^2+1)}]$ Use rule (2) above to obtain: $=\frac{1}{2}[\ln{(x-1)}+\ln{(x+1)}-\ln{(x^2+1)}]$
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