$$\forall x,y \in \mathbb{R} :f(x+y)=f(x)f(y) ,f(0)\neq0$$ Exponential function $f(x)=a^x $ works fine here .
My question :Is there any $\bf\color{red} {\text{other function}}$ exept of exponential functions work for this property ?
$\bf\text{I know like this question asked before}$ ,but My question is about uniqueness type of functions which work with this property.
On the other hand is there a proof that only exponential function work for that property ?
On
Let $g$ be any function satisfying $g(x + y) = g(x) + g(y)$. Then if we set $f = \exp g$,
$$f(x + y) = \exp\big(g(x) + g(y)\big) = f(x) f(y)$$
Functions $g$ are solutions to the Cauchy functional equation, of which there are many (typically constructed using choice), including anything of the form $ax$.
On
It's very easy to show that $f(0) = 1$ and if $f(1)= b$ then by induction $f(n)= b^n$ for any natural $n$ and that $f(\frac 1n) = \sqrt[n]b$, and so for all $q \in \mathbb Q$ that $f(q) = b^q$ ($b^{\frac nm} = (\sqrt[m]{b})^n$).
And so as $\mathbb Q $ is dense in $\mathbb R$ it follows that if $f$ if continuous, that $f(x) = b^x$.
But $f$ is not continuous it does not. For any irrational $k$ we could set $f(k)= c$ by any value and we could have and $f(qk) = f(k)^q$ for all rational $q$ but as $qk$ is not rational no contradiction need arise. ($f(r + qk) = b^rc^q \not \in \mathbb Q$ (unless $q=0$) for all $r,q \in \mathbb Q$.)
But we will have for any $w \in \mathbb R$ that for and $q_i\in \mathbb R$ that $f(\sum q_i*w^i) = \prod f(w^i)^{q_i}$.
From this one proves that $f(rm)=\left[f(m)\right]^r$ with $r$ rational.
It only has to be $a^x$ for each equivalence class of $m$ under the relation $x \sim y \iff xy \ne 0 \land \dfrac xy \in \Bbb Q$.
Partition the reals under this equivalence relation, and set the value of the function at each representative to your favourite real number, and you'll have a solution not in the form $a^x$ for all $x$.