If $X$ has density $f(x) = 2e^{−x}$ on $[0, \log 2]$ and $0$ elsewhere, what is $E[e^X]$?

Question

If $X$ has density $f(x) = 2e^{−x}$ on $[0, \log 2]$ and $0$ elsewhere, what is $E[e^X]$?

I thought that since $E[g(x)] = \int g(x)f(x)\, dx$, the answer would be $\int_0^{\log 2}e^x2e^{-x}\, dx = 2 \log 2$, but according to my answer key the answer is $\int_0^{\log 2}e^x2e^{-2x}\, dx$. Where did the $2$ in the exponent come from?

Answer 1

Since $\int_0^{\log 2} 2e^{-x} \mathop{dx}=1$ but $\int_0^{\log 2} 2e^{-2x} \mathop{dx} =3/4 \ne 1$, I suspect the answer key has a typo and that you are correct.