Verified Answer :

\(2e^{-t}-2te^{-t}=0\\\\2e^{-t}(1-t)=0\iff2e^{-t}=0\ or\ 1-t=0\\\\2e^{-t) > 0\ for\
each\ value\ of\ t\\\\1-t=0\iff t=1\)


Answer: t = 1.

\(2e^(^-^t^)-2te^(^-^t^)=0\) 

\(2e^-^t-2te^-^t=0\)     \(Simplify \ brackets\) 

\(2(\dfrac{1}{e^t}- \dfrac{t}{e^t})=0\)    \(Factor \ out \ the \ common \ term \ 2\) 

\(\dfrac{1}{e^t}- \dfrac{t}{e^t}=0\)  \(Divide \ both \ sides \ by \ 2\) 

\(\dfrac{1-t}{e^t} =0\)    \(Join \ denominators\) 

\(1-t=0\)   \(Multiply \ both \ sides \ by \ e^t\) 

\(-t=-1\)   \(Subtract \ 1 \ from \ both \ sides\) 

\(t=1\)    \(Multiply \ both \ sides \ by \ -1\)