Does a function that satisfies the equality $f(a+b) = f(a)f(b)$ have to be exponential? I understand the other way around, where if a function is exponential then it will satisfy the equality $f(a+b)=f(a)f(b)$. But is every function that satisfies that equality always exponential?
 A: First see that $f(0)$ is either 0 or 1. If $f(0)=0$, then for all $x\in\mathbb R$, $f(x)=f(0)f(x)=0$. In this case $f(x)=0$ a constant function.
Let's assume $f(0)=1$. See that for positive integer $n$, we have $f(nx)=f(x)^n$ which means $f(n)=f(1)^n$. Also see that:
$$
f(1)=f(n\frac 1n)=f(\frac 1n)^n\implies f(\frac 1n)=f(1)^{1/n}.
$$
Therefore for all positive rational numbers:
$$
f(\frac mn)=f(1)^{m/n}.
$$
If the function is continuous, then $f(x)=f(1)^x$ for all positive $x$. For negative $x$ see that:
$$
f(0)=f(x)f(-x)\implies f(x)=\frac{1}{f(-x)}.
$$
So in general $f(x)=a^x$ for some $a>0$. 

Without continuity, consider the  relation: $xRy$ if $\frac xy\in \mathbb Q$ (quotient group $\mathbb R/\mathbb Q$). This relation forms an equivalence class and partitions $\mathbb R$ to sets with leaders $z$. In each partition the function is exponential with base $f(z)$.
A: Suppose $f$


*

*is a real-valued function of a real variable, and

*is monotonic (i.e. either nowhere decreasing or nowhere increasing), and

*satisfies $f(a+b)=f(a)f(b)$ for all $a,b\in\mathbb R$.


Then $f$ is an exponential function (and in partcicular, $f$ is continuous).
However, a complex-valued function of a complex variable can be continuous and satisfy $f(a+b)=f(a)f(b)$ without being an exponential function.  Here is an example:
$$
f(x+iy) = 2^x(\cos y^\circ + i\sin y^\circ). \qquad (\text{for }x,y\text{ real, where }x^\circ\text{ means } x\text{ degrees}.)
$$
If that were exponential it would have base $2$, since its restriction to $y=0$ is $x\mapsto 2^x$.  However,
$$
2^{x+iy} = 2^x(\cos(y\log_e 2) + i\sin(y\log_e 2)) \qquad (\text{with radians, not degrees}).
$$
Arash's answer explains why in the real case it must be exponential (except that you should note that the hypothesis of continuity in Arash's answer can be weakened to monotonicity).
