Is this homomorphism identically zero? This question was motivated by trying to generalise the derivative of $\ln x$ to a homomorphism.

Let $f: \mathbb R^+ \to \mathbb R$ be such that $f(xy)=f(x)+f(y)$ for all positive real numbers $x$ and $y$.

We can show that $f(1)=0$ by considering $f(1\cdot y)=f(1)+f(y)$.

Assume that, for example, $f(2)=0$. We can show that $f(2^q)=0$ for all rational numbers $q$.

My first instinct was that $f(x)=0$ for all $x&gt;0$, but I'm probably wrong. 

QUESTION: If $f$ is continuous for all $x&gt;0$ then does it follow that $f(x) \equiv 0$? If not, then how about $f$ being differentiable?

EDIT: I know about the logarithm. This question was motivated by trying to find the derivative of a function that satisfies the "laws of logs". Finding the derivative of a log function requires that it be one-to-one. So I asked myself: what if it is not one-to-one? We must have $f(1)=0$, so I assumed that $f(2)=0$ and it implied that $f(2^q)=0$ for all $q \in \mathbb Q$. I then thought that perhaps, if $f$ is continuous then $f(x) \equiv 0$.



Proof that $f(2^q)=0$ for all $q \in \mathbb Q$.

We can show that $f(2^{1/k})=0$ for all positive integers $k$:

$$0 \ \ = \ \ f(2^1) \ \ = \ \ f(\underbrace{2^{1/k}\cdots 2^{1/k}}_{k \ \mathrm{times}}) \ \ = \  f(2^{1/k})+\cdots+f(2^{1/k}) \ \ = \ \ k\cdot f(2^{1/k})$$

It follows that $f(2^{-1/k})=0$ for all positive integers $k$:
$$0 \ \ = \ \ f(1) \ \ = \ \ f\left(\frac{1}{2^{1/k}} \cdot 2^{1/k}\right)\ \ = \ \ f\left(\frac{1}{2^{1/k}}\right)+f\left(2^{1/k}\right) \ \ = \ \ f\left(\frac{1}{2^{1/k}}\right)+0$$

We can show that $f(2^q)=0$ for all positive rational numbers $q=s/t$:

$$f(2^{s/t}) \ \ = \ \ f([2^{1/t}]^s) \ \ = \ \ f(\underbrace{2^{1/t} \cdots 2^{1/t}}_{s \ \mathrm{times}}) \ \ = \ \ s \cdot f(2^{1/t}) \ \ = \ \ s \cdot 0 \ \ = \ \ 0$$

It follows that $f(2^{-q})=0$ for all positive rational numbers $q=s/t$:
$$0 \ \ = \ \ f(1) \ \ = \ \ f\left(\frac{1}{2^{s/t}} \cdot 2^{s/t}\right)\ \ = \ \ f\left(\frac{1}{2^{s/t}}\right)+f\left(2^{s/t}\right) \ \ = \ \ f\left(\frac{1}{2^{s/t}}\right)+0$$

Hence $f(2^q)=0$ for all rational numbers $q$.
The endomorphisms of $\mathbb R$ are vast, they are already a lot if you consider the linear functions $\mathbb R \rightarrow \mathbb R$ when looking at $\mathbb R $ as a vector space over $\mathbb Q$, and each of these is a group homomorphism.

The homomorfisms from $\mathbb R^+ $ to $\mathbb R$ are also a lot, since if $\varphi$ is an isomorphicm on $\mathbb R$ you always have the morfism obtained via the following composition:

$\mathbb R^+ \overbrace{\rightarrow}^{\ln} \mathbb R 
\overbrace{\rightarrow}^\varphi \mathbb R $



Suppose $f$ is continuous and $f(2)=0$. We can compose $f$ with the function $g(x)=e^x$. This gives us a homomorphims $f\circ g :\mathbb R \rightarrow \mathbb R$ that vanishes at some point $\alpha\neq 0$.  Using the fact that $x$ is a homomorphism we can prove that $f(\alpha/n)=0$ for all $n$ and then prove that $f(m\alpha/n)=0$ also. Since this set is dense in $\mathbb R$ and the function is continuous we get that $f(x)=0$ for all $x$. Since $g$ is surjective it follows that $f$ is $0$ at every point.

In fact it suffices for $f$ to be measurable and vanish at some point, proving this is slightly harder and uses the Steinhaus theorem in measure theory.
The multiplicative group $(\Bbb{R}^{+}, \times)$ is isomorphic to the additive group $(\Bbb{R}, +)$ via the mapping $x \mapsto \ln(x)$. So your problem is equivalent to one about additive homomorphisms: functions $h : \Bbb{R} \to \Bbb{R}$ such that $h(x + y) = h(x) + h(y).$  If such an $h$ is continuous, then it must be given by $h(x) = \lambda x$ for some $\lambda \in \Bbb{R}$ (which you can see by thinking about the values $h(m/n)$ for integer $m$ and $n$ using the homomorphism property and then using continuity). This means that if your $f$ is continuous and has $f(2) = 0$ then $f(x) = 0$ for all $x$. If you don't assume continuity, then information about $f(2)$ will not determine $f(x)$ for all values of $x$.