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 - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 825: 21



Work Step by Step

Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the given equation, $ e^{-0.02t}=8 ,$ is \begin{array}{l}\require{cancel} \ln e^{-0.02t}=\ln 8 \\\\ -0.02t\ln e=\ln 8 \\\\ -0.02t(1)=\ln 8 \\\\ -0.02t=\ln 8 \\\\ t=\dfrac{\ln 8}{-0.02} \\\\ t\approx-103.972 .\end{array}
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