How to prove $x^ax^b = x^{a+b}$ I am looking for a proof of one of the exponent combination laws, namely the sum of powers. Here $x, a, b \in \mathbb R$ and $x > 0$. I thought about induction but since a,b are not only positive integers ($a, b \in \mathbb R$) that would not work out. Any suggestions?

It suffices to prove it for $x=e$ because then $x^ax^b=e^{a\ln x}e^{b\ln x}=e^{(a+b)\ln x}=x^{a+b}$.
$$e^ae^b=\sum_{n=0}^\infty\frac{a^n}{n!}\sum_{n=0}^\infty\frac{b^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{a^k}{k!}\frac{b^{n-k}}{(n-k)!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac1{n!}\binom nk a^kb^{n-k}=\sum_{n=0}^\infty\frac{(a+b)^n}{n!}=e^{a+b}$$
Here we use Cauchy's product formula, the series converge absolutely.
