How to prove $x^ax^b = x^{a+b}$

Question

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?

Answer 1

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.

Answer 2

I know of no proof simpler than the one already accepted, not to mention that it generalizes to $\mathbb{C}$. However it uses infinite sums and thus might not be available to students beginning an introductory course in Analysis, so I'll include a proof that can be read right after having defined $\mathbb{R}$.

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Let $x>0$.

Step 1: Show that for any natural $n>0$, there exists a unique $y>0$ such that $y^n=x$. This can be shown as in the first chapter of Rudin's Principles of Mathematical Analysis or as done in this post.

Step 2: Show that if $p,q,r,s$ are integers with $q,s>0$, and $\frac{p}{q}=\frac{r}{s}$, then $$(x^p)^{1/q}=(x^r)^{1/s}.$$ So that for a rational $p/q$ with $q>0$ we can define $x^{p/q}$ as $(x^p)^{1/q}$.

Step 3: Show that for rationals $a,b$ we have $x^ax^b=x^{a+b}$.

Step 4: For $x>1$, and $a\in \mathbb{R}$, let $x^a$ be the supremum of $\{ x^t:t\leq a\}.$ Show that for reals $a,b$ we have $x^ax^b=x^{a+b}$. Using basic properties of supremums i.e. $\sup(A)\sup(B)=\sup(AB)$ is helpful here.

Step 5: Let $(1/x)^a=1/x^a$ and show the results extends to all positive reals.