Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.3 Logarithmic Functions - 12.3 Exercise Set - Page 803: 61

Answer

$e^{-1.3863}=0.25$

Work Step by Step

Since $y=\log_bx$ is equivalent to $b^y=x,$ the given equation, $ \log_{e} 0.25=-1.3863 ,$ is equivalent to \begin{array}{l}\require{cancel} e^{-1.3863}=0.25 .\end{array}
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