Answer
$f(t)=u_0(t)+(2e^{-t}-1)u_{\ln 2}(t)$
Work Step by Step
Since $0 \leq t \lt \ln 2$, $f(t)=1$
then we have $f_1(t)=u_0(t)$
Since $t \geq \ln 2$, $f(t)=2e^{-t}$
then we have $f_2(t)=(2e^{-t}-1)u_{\ln 2}(t)$
Hence,
$f(t)=f_1(t)+f_2(t)\\
=u_0(t)+(2e^{-t}-1)u_{\ln 2}(t)$
