Is $e^x$ the only isomorphism between the groups $(mathbb{R},+)$ and $(mathbb{R}_{> 0},*)$?

If so, how might I be able to prove it?

EDIT: OK, thanks to many answers especially spin's and Micah's explanations. All of the answers were extremely helpful in understanding -- I have accepted Micah's because it seems the most complete, but all answers provide helpful additions/perspectives! I have tried to summarize:

$\phi$ is an isomorphism between the groups if and only if $\phi(x) = e^{f(x)}$ where $f$ is an isomorphism from $(\mathbb{R},+)$ back to $(\mathbb{R},+)$.

Of course there are lots of such $f$, especially when we take the Axiom of Choice.

However, it seems from the answers and Micah's link (Cauchy functional equation) that the only "nice" solutions are $f(x) = cx$ for a constant $c$. It seems that all others must be "highly pathological" (in fact $\{(x,f(x))\}$ must be dense in $\mathbb{R}^2$).

A remaining question is, how strong is the statement

All such isomorphisms have the form $e^{cx}$ for some $c \in \mathbb{R}$

or its negation? Or what is required for each to hold?

One answer seems to be that supposing the reals have the Baire property is sufficient to rule out other solutions (as is assuming every subset of the reals is measurable, assuming the Axiom of Determinacy, and it holds in Solovay's model). For more, see this question, this question, and this mathoverflow question.

Answer

Let $f:\mathbb{R} \to \mathbb{R}^+$ be an isomorphism. Then $g=\log f$ is an automorphism of $(\mathbb{R}, +)$: that is, it satisfies
$$
g(x+y)=g(x)+g(y) \, .
$$
This is Cauchy's functional equation. The only continuous (or even measurable) solutions are the trivial ones (which correspond to $f(x)=e^{cx}$), but there are also exotic solutions that require some version of the axiom of choice to construct — which would yield similarly exotic $f$s.