Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 342: 22


$ t= \frac{1}{3}\ln9$

Work Step by Step

We have $$ e^{2t+1}=9e^{1-t}\Longrightarrow \ln e^{2t+1}= \ln(9e^{1-t}) \\ \Longrightarrow {2t+1}= \ln9+\ln e^{1-t}\\ \Longrightarrow {2t+1}= \ln9+1-t\\ \Longrightarrow 3t= \ln9\Longrightarrow t= \frac{1}{3}\ln9. $$
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