Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 27

Answer

$ln(z)+2ln(z-1)$

Work Step by Step

Use the rule that $log(ab)=log(a)+log(b)$ to simplify the logarithm to $ln(z)+ln((z-1)^{2})$. Finally, use the rule that $log(x^{n})=nlog(x)$ to reduce the $ln((z-1)^{2})$ and reach $ln(z)+2ln(z-1).$
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