Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set: 22

Answer

${\frac{1}{5}}$

Work Step by Step

A natural logarithm is a logarithm to base $e$. As shown on page 582, when a natural logarithm is written in the form $ln (x)$, this is equivalent to the expression $log_{e}x$. Therefore, $ln \sqrt[5]e=ln (e^{\frac{1}{5}})=log_{e}e^{\frac{1}{5}}=\frac{1}{5}$. We know this, because $(e)^{\frac{1}{5}}=e^{\frac{1}{5}}$.
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