$ \sum_{k=0}^\infty (-1)^k \frac{(\ln 4)^{k}}{k!} $

Question

I'm taking single variable calculus on Coursera platform and asked :

Calculate $$ \sum_{k=0}^\infty (-1)^k \frac{(\ln 4)^{k}}{k!} $$

The natural logarithm of a number is its logarithm to base $e$ but how can this be calculated with above formula ?

Update :

substituting $x=-a$ , $a=\ln4$ in $\sum_{k=0}^{\infty}\frac{x^k}{k!}=e^x$

gives :

$$ \sum_{k=0}^\infty (-1)^{k}\frac{(\ln 4)^{k}}{k!}=\sum_{k=0}^\infty \frac{(-\ln 4)^{k}}{k!} =e^{-\ln4}=\frac14 $$

Answer 1

Hint: $$ \sum_{k=0}^{\infty}\frac{x^k}{k!}=e^x $$

What if $x=-a$?

And what if $a=\ln4$?