When a real valued function is multiple of the natural logarithm I am trying to prove the next assertion:

If $f:]0,\infty[ \longrightarrow \mathbb{R}$ is a continuous function
  with the following property:$$\forall x,y\in ]0,\infty[ \quad f(x\cdot y)=f(x)+f(y)$$
  there is $k\in \mathbb{R}$ such that $f(x)=k \log(x)$.

I was trying to prove that $k=f'(1)$, but I was not able to prove that $f'(1)$ exists.
 A: Hint: try considering $g:(0,+\infty)\to\Bbb R$, $g(x)=e^{f(x)}$, then $g(xy)=g(x)g(y)$ and $g$ is continuous. It should now be easy to conclude on the nature of $g$.
A: The best way to do this is to consider why $f$ must be the inverse of an exponential function. That is, we want to show that $f(e^x)=kx$ for all $x>0$, where $k=f(e)$. We do this in stages.
Firstly, note that $f(1)=0$ since $f(1)=f(1)+f(1)$ and if $f(e^n)=kn$ then $f(e^{n+1})=f(e^n)+f(e)=kn+k=k(n+1)$, so by induction $f(e^n)=kn$ for all $n\in\mathbb{N}$.
Next, look at negative integers: since $f(e^{-n})+f(e^n)=f(1)=0$, it follows that $f(e^{-n})=-f(e^n)=-kn$.
Now we look at fractional powers. Note that $f(x^n)=nf(x)$ for all $x>0,n\in\mathbb{N}$, so since $(e^{m/n})^n=e^m$, it follows that $km=f(e^m)=nf(e^{m/n})$. Hence $f(x)=kx$ for all $x\in\mathbb{Q}_+$, so by the density of $\mathbb{Q}_+\subset\mathbb{R}_+$ we get the result.
A: Here's a alternative solution. 
Since $f(1)=f(1)+f(1)=2f(1)$, we deduce that $f(1)=0$, and for every $x\in (0,\infty)$ and every $n\in \mathbb{N}$ we have
$f(x^n)=f(x^{n-1})+f(x)$, and so by induction we have $f(x^n)=nf(x)$. It follows that $f(x^{-n})=f(x^{-n})+f(x^n)-f(x^n)=f(1)-nf(x)=-nf(x)$. 
Hence 
$$\tag{1}
f(x^n)=nf(x) \quad \forall x\in (0,\infty), n\in \mathbb{Z}.
$$
If $n\in \mathbb{Z}\setminus\{0\}$, Thanks to (1), we have for every $x\in (0,\infty)$:
$$
f(x^{1/n})=\frac1n\cdot nf(x^{1/n})=\frac1nf\left((x^{1/n})^n\right)=\frac1nf(x)
$$
i.e.
$$\tag{2}
f(x^{1/n})=\frac1nf(x)\quad \forall x\in (0,\infty), n\in \mathbb{Z}\setminus\{0\}.
$$
Let $r=p/q\in \mathbb{Q}$, and let $x\in (0,\infty)$. Combining (1) and (2), we have:
$$
f(x^r)=f\left((x^{1/q})^p\right)=pf(x^{1/q}))=\frac{p}{q}f(x)=rf(x)
$$
Let $\alpha\in \mathbb{R}, x\in (0,\infty)$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there is a convergent sequence $\{r_n\}$ in $\mathbb{Q}$ whose limit is $\alpha$. Using the fact that $f$ is continuous we have
$$
f(x^\alpha)=f(\lim_n x^{r_n})=\lim_nf(x^{r_n})=\lim_nr_nf(x)=\alpha f(x),
$$
i.e.
$$\tag{3}
f(x^\alpha)=\alpha f(x)\quad \forall x\in (0,\infty), \alpha\in \mathbb{R}.
$$
Since $x=10^{\log x}$ for every $x\in (0,\infty)$, thanks to (3) we have:
$$
f(x)=f\left(10^{\log x}\right)=f(10)\log x=k\log x \quad \forall x\in (0,\infty),
$$
with $k=f(10)$.
