If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t 
Let  $f(xy) =f(x)f(y)$ for all $x,y\geq 0$. Show that $f(x) = x^p$ for some $p$. 

I am not very experienced with proof. If we let $g(x)=\log (f(x))$ then this is the same as  $g(xy) = g(x) + g(y)$ 
I looked up the hint and it says let $g(x) = \log f(a^x) $ 
The wikipedia page for functional equations only states the form of the solutions without proof.
Attempt 
Using the hint (which was like pulling a rabbit out of the hat)
Restricting the codomain $f:(0,+\infty)\rightarrow (0,+\infty)$
so that we can define the real function $g(x) =  \log f(a^x)$ and we
have $$g(x+y) = g(x)+ g(y)$$ 
i.e $g(x) = xg(1)$ as $g(x)$ is continuous (assuming $f$ is).
Letting $\log_a f(a) = p$ we get $f(a^x) =a^p $. I do not have a rigorous argument but I think I can conclude that $f(x) = x^p$ (please fill any holes or unspecified assumptions) Different solutions are invited

So, we assume $f$ is continuous. Letting $g(x) = \ln(f(a^x))$, we get
$$
\begin{align*}
g(x+y) &= \ln(f(a^{x+y})) = \ln(f(a^xa^y)) = \ln(f(a^x)f(a^y))\\
 &= \log(f(a^x)) + \ln(f(a^y))\\
 &= g(x)+g(y).
\end{align*}$$
So $g$ satisfies the Cauchy functional equation; if you assume $f$ is continuous, then so is $g$, hence $g(x) = xg(1)$ for all $x\gt 0$.
Since $g(1) = \ln(f(a))$, we have
$$f(a^x) = e^{g(x)} = e^{g(1)x} = (e^{x})^{g(1)}.$$
Given $r\in \mathbb{R}$, $r\gt 0$, we have $r = a^{\log_a(r)}$, hence
$$\begin{align*}
f(r) &= f\left(a^{\log_a(r)}\right)\\
     &= \left(e^{\log_a(r)}\right)^{g(1)}\\
     &= \left(e^{\ln(r)/\ln(a)}\right)^{g(1)}\\
     &= \left(e^{\ln(r)}\right)^{g(1)/\ln(a)}\\
     &= r^{g(1)/\ln(a)},
\end{align*}$$
where we have used the change-of-base formula for the logarithm,
$$\log_a(r) = \frac{\ln r}{\ln a}.$$
Finally, since $g(1) = \ln(f(a))$, we have
$$f(r) = r^{\ln(f(a))/\ln(a)}.$$
As this works for any positive $a$, $a\neq 1$, taking $a=e$ we get
$$f(r) = r^{\ln(f(e))}.$$
