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It is certainly true that the only $differentiable$ homomorphisms are exponential maps. To see this let $f$ be a differentiable homomorphism, and note that $f(0) = 1$. Write $f'(0) = a$, so $\lim_{t \to 0} (f(t) - 1)/t = a$. Now we compute $f'(x)$ for arbitrary $x$. We have $$ f'(x) = \lim_{t \to 0} \frac{f(x+t) - f(x)}{t}, $$ and $f(x+t) = f(x)f(t)$ since $f$ is a homomorphism. This yields $$ f'(x) = \lim_{t \to 0} f(x)\frac{f(t) - 1}{t} = a f(x). $$ The unique solution to this differential equation with initial condition $f(0) = 1$ is $f(x) = \exp(ax)$.