**Question:**


what can be said about the existence of functions  

\begin{align}
f:x\mapsto f(x)&\implies x+c\mapsto \gamma(c)f(x)\\ f(x)\ne f(y)&\iff\frac{f(x)}{\gamma(x)}\ne\frac{f(y)}{\gamma(y)}
\end{align}


These functions would generalize the functional equation $e^{x+y}=e^xe^y$


I am especially interested in the case $x,f(x)\,\in\,\mathbb{R}$, but also appreciate answers related to different types of functions and arguments.



Non-exponential functions $f(x)$ satisfying $f(x+c)=\gamma(c)f(x)$
Assume that $f(x)$ and $g(x)$ are real functions satisfying the relationship $f(a+b)=f(a)g(b)$. Then one must have $f(x)=Kg(x)$ for some constant $K$.


Proof:
Observe that $f(0)=f(0+0)=f(0)g(0)$. So we have that either $f(0)=0$ or $g(0)=1$.


Consider the case where $g(0)=1$. Then we have that $f(a+0)=f(a)g(0)$ but $f(a+0)=f(0+a) = f(0)g(a)$. So $f(a)g(0)=f(0)g(a)$ and since $g(0)=1$ we have $f(a)=f(0)g(a)$ for all $a$.


Now, consider the case $f(0)=0.$ Then we have $f(x-x)=f(x)g(-x)=0$ for all $x$. But if $f(x)$ is zero for any $x$, then we must have $f(x)$ is always zero. So we must have $g(x)=0$ for all $x$ which also implies that $f(x)=0$.


So our only possible non-trivial cases are $f(x)=Kg(x)$ for some non-zero $K$.


