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

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

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