College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.4 - Evaluating Logarithms and the Change-of-Base Theorem - 4.4 Exercises: 45

Answer

$1.6$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of logarithms to evaluate the given expression, $ \ln e^{1.6} .$ $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} 1.6\ln e .\end{array} Since $\ln e=1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 1.6(1) \\\\= 1.6 .\end{array}
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