Let $f: \mathbb{R} \to \mathbb{R}$ such that $f(a+b) = f(a)f(b)$ and $f\neq 0$. Can I deduce from this that $f$ is an exponential function i.e. does there exist $0 < c \in \mathbb{R}$ such that $f = c^{x}$.
These are my ideas so far:
From the definition we get $f(a) = f(a+0) = f(a)f(0)$ for all $a$. Using $f \neq 0$ we get $f(0) = 1$.
We also have $f(n) =f(1+\ldots + 1) = f(1) \cdots f(1) = f(1)^n$ so if $f$ was of the form $f(x) = c^{x}$ then we would need $c = f(1)$ but I don't know if this really is the only option, or if there are functions with this property that are not exponentials.