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

Solution 1:

Once you have an isomorphism, there are plenty. Take two Hamel bases of $\mathbf{R}$, say $(b_i)$ and $(c_i)$, and define $f$ by $$ f(b_i) = e^{c_i}, $$ extending it by $\mathbf{Q}$-linearity to all of $\mathbf{R}$.

PS Of course I am implicitly using here the argument of @Micah and @spin.

Solution 2:

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.

Solution 3:

$f:\mathbb{R}\rightarrow\mathbb{R}^{+} \text{ satisfying } f(x+y)=f(x)f(y)$, $f$ continuous, then $f(x)=e^{cx}$

Try to solve this problem Any Lie Group Homomorphism from $\mathbb{R}\rightarrow S^1$ is of the form $e^{iax}$ for some $a\in\mathbb{R}$ and every such homomorphism is smooth.