Answer
$2$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
As we are given that $x=\ln 2$
Then, we have
$e^{x}=e^{(\ln 2)}=2$
