I come from a physics background and in classical mechanics, we construct a Hamiltonian function whose partial derivatives generates a vector field, two independent systems are assigned a total Hamiltonian which is the algebraic sum of the Hamiltonians for the isolated systems.
In statistical mechanics, Liouville's theorem requires that in equilibrium, the probability that a system's dynamical variables have specific values is a function of the value Hamiltonian for those specific values. So the standard probability theory argument that the probability to measure state a in system A and state b in system B for independent systems A and B is the product of the individual probabilities: $Pr(a \text{ & } b) = Pr(a)*Pr(b)$. This is essentially a quick derivation of the maxwell-boltzmann distribution.
I've seen this argument about probability theory in several places, and can easily see that the exponential satisfies the requirement.
So I think my question is essentially this, does $$ f(a+b) = f(a)*f(b) \\ f(x) > 0 \text{ for all real x} \\ $$
uniquely specify that $f(x)=Ae^{cx}$? Or is there some other consideration that I must be overlooking?
If $f$ is a differentiable function, taking a derivative with respect to $b$ means that $$f'(a+b)=f(a)f'(b).$$ Now substitute $b=0$ to get the differential equation $f'(a)=f(a) f'(0)$ which has the solution $f(x) = A e^{f'(0) x}$. Thus $f'(x) = f'(0) A e^{f'(0) x}$. Solving for $f'(0)$ gives $f'(0)=Af'(0)$ which is only true if $A=1$ or $f'(0)=0$. In the latter case we end up with the constant function $f=1$ (since $A=A^2$). Otherwise, the solution is $f(x) = A e^{c x}$ where choosing $c$ chooses $f'(0)$.