Powerseries of $b^x$

Question

I read on wikipedia that the exponential function $e^x$ can be written(defined) in this form

$\displaystyle e^x:=\sum_{n=0}^{\infty}({1\over n!})x^n$

So my question was if its is then possible to express $b^x$ as powerseries using the fact that $b^x=e^{\ln(b)x}$

Applying in a naive way some algebraic rules I get this

$\displaystyle b^x=e^{\ln(b)x}=\sum_{n=0}^{\infty}({1\over n!})({\ln(b)x})^n=\sum_{n=0}^{\infty}{\ln(b)^n\over n!}{x}^n$

Anyways even if it seems obvious to me I feel that my knowledge of what is a powerseries and how it works is too poor to make me sure. For example searching for "powerseries of b^x" a Wolfram Alpha it gives me the following eqaution

$\displaystyle b^x=\sum_{n=0}^{\infty}{\ln^n(b)\over n!}{x}^n$

So it is $\ln^n(b)$(iteration) or $\ln(x)^n$(power)?

PS:I've also asked because I couldn't find a closed form for $b^x$ at wikipedia.

Answer 1

In this context $\ln(b)^n=\ln^n(b)$ $$ \mathrm{e}^{ax} = \sum_{i=0}^{\infty}\frac{(ax)^n}{n!} $$ As you have already shown with $a = \ln b$

Answer 2

Wolfram Alpha is using $\ln^n b$ to mean the $n$th power of $\ln b$. This is an abuse, but a standard abuse, of notation. Kind of like $\sin^2 x + \cos^2 x = 1$.