Functions such that $f(ax)=f(x)+f(a)$ I thought of this question somewhat randomly on a walk, and have discussed it with another friend of mine (we both have pure mathematics degrees). We have made some headway, and we think we have generated a proof, but we would appreciate any additional insight and proof verification.

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(ax)=f(x)+f(a)$. How much can we say about the behavior of $f(x)$? What additional restrictions, if any, allow us to show $f(x)=\log_b(x)$?

We conjecture that, with the restriction that $f$ is not $0$ everywhere, $f$ must be the logarithm. We have, by the log rules,
$$\log_b(|ax|)=\log_b(|x|)+\log_b(|a|)$$
So the logarithm is indeed one such $f$. I will give what we have been able to show about $f$ below, followed by our general proof. If you can offer any insight into this problem, we would greatly appreciate it.
Edit: Turns out this can be shown fairly easily by considering the Cauchy Functional Equation and showing $g(x)=f(e^x)=cx$ for some $c$. The proof given below does not use this fact.
From here on, we assume $f$ is not trivial.

Result 1: $f(1)=0$
We note that 
$$f(a)=f(a\cdot1)=f(1)+f(a)$$
which shows $f(1)=0$.

Result 2: $f(0)$ is not defined
We see that 
$$f(0)=f(a\cdot0)=f(0)+f(a)$$
which implies $f(a)=0$ for all $a$. However, we have assumed $f(x)\neq0$ for some $x$, giving a contradiction. Thus $f(0)$ must not be defined. So, we redefine $f$ as $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$.

Result 3: $f(-x)=f(x)$
If we allow $x<0$, we then have
$$0=f(1)=f(-1\cdot-1)=2f(-1)$$
Showing that $f(-1)=0$ as well. Using this result, we have
$$f(-a)=f(-1)+f(a)=f(a)$$

Our Proof:
From here on, we use results from elementary abstract algebra and analysis. We realized that the condition on $f$ is that of a group homomorphism $\varphi:(\mathbb{R}\setminus\{0\},*)\to(\mathbb{R},+)$ where 
$$\varphi(xy)=\varphi(x)+\varphi(y)$$
Similarly, you can show the above results for $\varphi$. We noted that if we also define the continuous group homomorphism $\psi:(\mathbb{R},+)\to(\mathbb{R}\setminus\{0\},*)$ where
$$\psi(x+y)=\psi(x)\psi(y)$$
$\psi$ and $\varphi$ seem like they could possibly be inverses with some additional restrictions. We see that 
$$\psi(x)=\psi(0+x)=\psi(0)\psi(x)=1$$
So that $\psi(0)=1$. We can then say for any integer $n\geq0$,
$$\psi(n)=\psi(1+1+1+...+1)=\psi(1)\cdot\psi(1)\cdot\cdot\cdot\psi(1)=\psi(1)^n$$
Note, $\psi(1)<0$ may give us complex results, so we restrict $\psi(1)>0$. For any integer $n<0$,
$$\psi(n)=\psi(-1-1+...-1)=\psi(-1)^{|n|}=(\psi(1)^{-1})^{|n|}=\psi(1)^{-|n|}=\psi(1)^n$$
Then, for any rational number $\frac{p}{q}$,
$$\psi(\frac{p}{q})=\psi(\frac{1}{q})^p=\psi(q^{-1})^p=(\psi(q)^{-1})^p=\psi(1)^{\frac{p}{q}}$$
Let $x\in\mathbb{R}\setminus\mathbb{Q}$. Since the rationals are dense in the reals, we can find rational $\frac{p}{q}<x<\frac{r}{s}$ arbitrarily close to $x$. Note that $\psi$ is strictly increasing over the rationals, and thus
$$\psi(1)^{\frac{p}{q}}<\psi(x)<\psi(1)^{\frac{r}{s}}$$
Since $\psi$ is continuous, this implies $\psi(x)=\psi(1)^x$ for all $x\in\mathbb{R}$. Therefore,
$$\psi(x)=\psi(1)^x$$
This shows that $\psi$ must be the exponential function.  Note that it is completely characterized by its value at $1$.  Noting that $\varphi$ seems like it should be related to the inverse of $\psi$, we would expect $\varphi(x)=\log_b(x)$. We note that for integer $n\geq0$ and real $a>0$,
$$\varphi(a^n)=\varphi(a)+\varphi(a)+...+\varphi(a)=n\varphi(a)$$
For $n<0$, we have
$$\varphi(a^n)=\varphi(\frac{1}{a})+...+\varphi(\frac{1}{a})=|n|\varphi(a^{-1})=-|n|\varphi{a}=n\varphi(a)$$
The rational case is slightly trickier. We note that $a^{\frac{n}{n}}=a$, and thus
$$\varphi(a)=\varphi(a^{\frac{n}{n}})=n\varphi(a^{\frac{1}{n}})$$
And thus $\varphi(a^{\frac{1}{n}})=\frac{1}{n}\varphi(a)$. Therefore, for any rational $\frac{p}{q}$
$$\varphi(a^{\frac{p}{q}})=p\varphi(a^{\frac{1}{q}})=\frac{p}{q}\varphi(a)$$
Using the same argument as before, we can extend this to all $x\in\mathbb{R}$. Therefore,
$$\varphi(a^x)=x\varphi(a)$$
Combining these, we have
$$\varphi(\psi(x))=\varphi(\psi(1)^x)=x\varphi(\psi(1))$$
Thus, $\varphi$ and $\psi$ are inverses up to multiplication by a constant, which confirms that $\psi$ must be the logarithm. If we restrict $\varphi(\psi(1))=1$, these are exactly inverses. If $\psi(1)>0$, we must have $\varphi_{\psi(1)}(x)=\log_{\psi(1)}(x)$. This proof requires that $x>0$. However, we showed earlier that $\varphi(-x)=\varphi(x)$, and for positive $x$, $\varphi_b(x)=\log_b(x)$. To make this function even, we modify it as 
$$\varphi_b(x)=\log_b|x|$$
and this is the only possible continuous solution for $\varphi$ defined on all of $\mathbb{R}\setminus\{0\}$. In other words, $f$ must be the logarithm. We have shown the logarithm rules are unique to logarithms!

One thing that concerns us is the restriction that $f$ is continuous. Are there discontinuous $f$ that satisfy this property? Please let us know if we made any invalid assumptions somewhere, or if our statements about the uniqueness of these functions is incorrect. We would also appreciate any alternate proofs you may have.
 A: This function is equivalent to the known as the cauchy functional equation. "Most" of the solutions are in fact discontinuous.
However all noncontinuous functions have a graph that is dense. So it suffices to show, that the graph is not dense (eg. f is monotone).
The original cauchy functional equation is $g(x+y) = g(x) +g(y)$. Let $g_0$ be a solution of this functional equation, then $h(x) = g_0(\log(x))$ is a solution to your equation. This is why (sometimes) your equation is also called the cauchy functional equation, because it's equivalent.
Futhermore the equations $f(x+y) = f(x)\cdot f(y)$ and $f(x\cdot y) = f(x)\cdot f(y)$ are equivalent aswell. 
A: Now, you see
$f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$
$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x(1+\frac{h}{x}))-f(x)}{h}}$
Now since, 
$ f(ax)=f(x)+f(a)$
$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(x)+f(1+\frac{h}{x})-f(x)}{h}} $
$ \implies f^{'}(x)= \lim \limits_{h \to 0} {\frac{f(1+\frac{h}{x})}{h}} $
Now, since we've considered $ f$ to be a derivable function, the limit is of the $ \frac{0}{0}$ form.
Using L'Hospital's rule and solving,
$ f^{'}(x)=f^{'}(1)\frac{1}{x}$
So, taking the anti-derivative and since $ f^{'}(1)$ is constant,
$ f(x)=c\ln |x|+\ln k$
Or,
$ f(x)=c\ln |kx|$
Now, since $ f(1)=0$, implies either c=0 or k=1.
And, since for c=0, doesn't satisfy the original condition, k=1;
So, $ f(x)=c\ln |x|$
