show that $f(x)=c\log x $ for some $c$ Let, $f$: $\mathbb{R^+}$$\rightarrow$$\mathbb{R}$ be a continuous function satisfying $f(xy)=f(x)+f(y)$. Prove that, $f(x)=c\log x$ for some $c>0$.
 A: It is obvious that the (continuous) function $\theta:\mathbb{R}^+\to \mathbb{R},\ \theta(x)=0$ satisfies the functional equation
$$\tag{*}
f(xy)=f(x)+f(y).
$$
Let $f \not\equiv \theta$ be a solution of (*). Then we have
$$
f(1)=f(1)+f(1),
$$
i.e.
$$\tag{1}
f(1)=0
$$
By induction we have
$$\tag{2}
f(x^n)=nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{N}.
$$
Thanks to (1), we have
$$\tag{3}
f(x^{-1})=f(x^{-1})+f(x)-f(x)=f(x^{-1}x)-f(x)=f(1)-f(x)=-f(x)\quad \forall x \in \mathbb{R}^+.
$$
Combining (2) and (3), we deduce that
$$\tag{4}
f(x^n)=nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{Z}.
$$
Thanks to (4), we have
$$
f(x^{1/n})=\frac1nf((x^{1/n})^n)=\frac1nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{N}.
$$
Thus
$$
f(x^{m/n})=\frac{m}{n}f(x)\quad \forall x \in \mathbb{R}^+,\,m\in \mathbb{Z},\, n\in \mathbb{N},
$$
in other words we have
$$
f(x^q)=qf(x)\quad \forall x \in \mathbb{R}^+,\,q\in \mathbb{Q}.
$$
Since $f$ is continuous, and $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that
$$
f(x^\alpha)=\alpha f(x)\quad \forall x \in \mathbb{R}^+,\,\alpha\in \mathbb{R}.
$$
Since $f\not\equiv\theta$, there is some $z \in \mathbb{R}^+\setminus\{1\}$ such that $f(z)\ne 0$. W.l.o.g. assume $1<z<e$ and $f(z)>0$ (otherwise we replace $z$ by $z^{-1}$). Let $s=\frac{1}{\ln z} \in \mathbb{R}^+$. Then, $z^s=e$ and for every $x \in \mathbb{R}^+$ we have
$$
f(x)=f(e^{\ln x})=f(e)\ln x=f(z^s)\ln x=sf(z)\ln x=c\ln x,
$$
where $c=s\ln z$. 
Remark: It is obvious that the constant $c$ need not be positive.
A: First show that $f\left(x^{\frac{m}{n}}\right) = \frac{m}{n}f(x)$ for any rational number $\frac{m}{n}$. Then use continuity to get $f(x^\alpha) = \alpha f(x)$ for all $\alpha \in \mathbb R$.
Now note that $f(1) = 0$. Suppose $f(k) = 0$ for some $k > 1$. Then for all $x$ we have $f(x) = f(k^\alpha) = \alpha f(k) = 0$ for some $\alpha$. Hence $f(x) = c\log(x)$ with constant $c = 0$.
Otherwise suppose, without loss of generality, that $f(k) > 0$ for some $k > 1$ (if $f(k) < 0$, then consider the function $-f$ for the following). Then there is some $\alpha > 0$ so that $f(k^\alpha) = \alpha f(k) = 1$. Let $b = k^\alpha$.
Then for any $x$ we have $x = b^\alpha$ where $\alpha = \frac{\log(x)}{\log(b)}$. So $f(x) = f(b^\alpha) = \alpha f(b) = \alpha = \frac{\log(x)}{\log(b)}$. Hence $f(x) = c\log(x)$ with constant $c = \frac{1}{\log(b)}$.
