If $a>1$. Prove that, for any real numbers $x, y:a^{x+y} = a^x a^y$ If $a>1$. Prove that, for any real numbers $x, y$:$$a^{x+y} = a^x a^y$$ $$(a^x)^y = a^{xy}$$ 
I feel that if $x,y$ were natural numbers the proof could be by induction but in our case how it can be done? Could anyone help me please?
EDIT:






 A: Here is a method if one does not know the proper definition of the exponential and logarithm functions. One first defines $a^r$ for $r\in \mathbf{Q}$ so that if $r = m/n$ then
$$a^{r} = \left(a^{n}\right)^{\frac{1}{m}} = \sqrt[m]{a^n}.$$
Since the $m$th root of $a$ is well defined this definition is valid. It is then easy to see that $a^r$ is increasing for $r\in \mathbf{Q}$ and therefore the definition of $a^x$ for $x\in \mathbf{R}$ is
$$\sup_{\substack{r<x\\r\in \mathbf{Q}}}a^r.$$
If you can show that $a^{r+q} = a^ra^q$ for any $r,q\in \mathbf{Q}$ then you can show that $a^{x+y}=a^xa^y$ for $x,y\in \mathbf{R}$. Here is how: Pick $r,q\in \mathbf{Q}$ such that $q<x$ and $r<y$ but so that $a^{r+q}>a^{x+y}-\varepsilon$ then
$$a^{x+y}-\varepsilon<a^{q+r} = a^qa^r\leq a^xa^y$$
and since $\varepsilon>0$ is arbitrary we find that $a^{x+y}\leq a^xa^y$. Now pick  $q<x$ and $r<y$ so that $a^q>a^x-\varepsilon$ and $a^r>a^y-\varepsilon$ and also let $r'$ be a rational greater than $x$ and $y$ then
$$a^{x}a^{y}-\varepsilon C<a^{x}a^{y}-\varepsilon(2a^{r'}-\varepsilon)<a^{x}a^{y}-\varepsilon(a^x+a^{y}-\varepsilon)=(a^x-\varepsilon)(a^y-\varepsilon)<a^{q}a^{r}=a^{q+r}\leq a^{x+y}$$
with $C = 2a^{r'}$. It is clear that this implies that $a^{x}a^{y}\leq a^{x+y}$ and therefore we may conclude that $a^{x+y} = a^{x}a^{y}$

To show that $a^{r+q} = a^{r}a^{q}$ the hint is to use that if $x,y$ are real then $(xy)^n = x^ny^n$ if $n\in \mathbf{N}$ together with the uniqueness of $n$th roots that is if $x,y>0$ then 
$$x^n = y^n$$
implies that $x = y$ if $n$ is a positive integer.
Specifically, say $r = n_1/m_1$ and $q = n_2/m_2$ then 
$$\left(a^{q+r}\right)^{m_1m_2} = \left(a^{\frac{n_1m_2+n_2m_1}{m_1m_2}}\right)^{m_1m_2} = a^{n_1m_2+n_2m_1} = \left(a^{\frac{n_1}{m_1}}\right)^{m_1m_2}\left(a^{\frac{n_2}{m_2}}\right)^{m_1m_2} = (a^{q}a^{r})^{m_1m_2}.$$
Now consider uniqueness of roots

Concerning the identity $(a^{x})^{y} = a^{xy}$ we first show that $(a^{x})^{r} = a^{xr}$ for $r$ rational. If $r = n/m$ then by what we showed above
$$\left((a^{x})^{r}\right)^{m} = (a^{x})^{n} = \underbrace{a^xa^x\cdots a^x}_{n\text{ times}} = a^{nx}.$$
Also
$$(a^{rx})^{m} = a^{\tfrac{n}{m}x+\cdots +\tfrac{n}{m}x} = a^{nx}$$
by uniqueness of roots we have thus shown that $(a^{x})^{r} = a^{rx}$. Now to complete the proof let $\varepsilon>0$ be given and pick $r<y$ such that 
$$(a^x)^y-\varepsilon<(a^{x})^{r} = a^{xr}\leq a^{xy}.$$
This shows that $(a^{x})^{y}\leq a^{xy}$. I leave it to you to try to prove the reverse inequality but write in the comments if you need help with this.
A: With the usual definition (as already written by Theo in the comments), the proof of the first identity goes as follows $$a^{x+y}=\exp((x+y)\ln(a))=\exp(x\ln(a))\exp(y\ln(a))=a^xa^b.$$
Here we used that $\exp(x+y)=\exp(x)\exp(y)$.
Now try to figure out the second one on your own. If you can't manage, I'll add a solution here.
A: For  $a^{x+y}:$
$$\begin{align}
a^{x+y} &= e^{(x+y)\ln a}\\
        &= e^{x\ln a+y \ln b}\\
        &= exp (x\ln a+y\ln b)\\
        &= exp(x\ln a)+ exp(y\ln b)\\
        &= e^{x\ln a}e^{y\ln a}\\
        &= a^xa^y
 \end{align}$$
For  $(a^x)^y:$
$$\begin{align}
(a^x)^y &= e^{y\ln(a^x)} \\
        &= e^{xy\ln a}\\
        &= a^{xy}
         \end{align}$$
