Rudin in Principles of Mathematical Analysis defines $b^x$ here as:
If $b>1$ and $x$ is any real, then $b^x=\sup B(x)$ where $B(x)$ is the set of all real numbers $b^t$ where $t$ is rational and $t\le x$.
I want to show that 1)$(b^x)^y=(b)^{xy}$ and 2)$a^xb^x=(ab)^x$.
Edit: My attemt:
1) (Assuming that $b^x>1$) Let $A=\{(b^x)^m:m\in \Bbb Q,m\le y \}$ and $B=\{b^t:t\in \Bbb Q,t\le xy \}$. Suppose $\sup A< \sup B$, then there is a rational $t_0$ with $t_0\le xy$ such that $b^{t_0}>(b^x)^m$ for all $m$ where $m$ is a rational with $m\le y$. (I am not quite clear here:) Since $b^x=B(x)$ there exists a rational $p\le x$ such that $b^{t_0}>(b^p)^m$, but since $pm\le xy$, we see that $b^{pm}\in B$, hence $b^{t_0}>(b^x)^m$ does not hold for all $m$. I can't come up with a contradiction if I assume $\sup B< \sup A$.
2) We have to show that $\sup A(x)\cdot \sup B(x)=\sup C(x)$ whre $C(x)=\{(ab)^t: t\mathrm{\, is\, a\, rational\, and \,} t\le x\}$. We will call $A(x)$ as $A$ and so on. If $p\in C$, then $p=(ab)^t=a^tb^t$ for some $t$ where $t$ rational and $t\le x$, so, for that $t$, $a^t\in A$ and $b^t\in B$, so $\sup A\cdot\sup B$ is an upper bound for $C$.
Now take $0<c<\sup A\cdot\sup B$. Hence $\frac{c}{\sup A}<\sup B$, so taking $m=\frac 12\left(\frac{c}{\sup A}+\sup B\right)$, we see that $\frac{c}{\sup A}<m<\sup B$, hence there is an $a^k\in A$ such that $\frac cm<a^k$ and $b^l\in B$ such that $m<b^l$. So, $\frac cm\cdot m<a^kb^l$. There exists a rational $v\le x$ such that $a^kb^l\le a^vb^v$ (I am not sure how to prove the existence of this $v$) so, $c<(ab)^v\in C$, hence $c$ is not an upper bound of $C$.