Both groups $\mathbb{R}^* / \mu(\mathbb{R})$ and $\mathbb{C}^* / \mu(\mathbb{C})$ (where $\mu$ is the group of roots of unity) are uniquely divisible, and thus are vector spaces over $\mathbb{Q}$.
Since the positive reals are closed under multiplication, it's easy to see that
$$ \mathbb{R}^* \cong \mu(\mathbb{R}) \oplus \mathbb{R}^* / \mu(\mathbb{R}) $$
I doubt that $\mu(\mathbb{C})$ is a direct summand of $\mathbb{C}^*$. However, this is still enough to give a positive answer to the question: we can, using the axiom of choice, construct a group homomorphism
$$ \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C})
\cong \mathbb{Q}^{\mathfrak{c}} \cong \mathbb{R}^* / \mu(\mathbb{R}) \to \mathbb{R}^*$$
Interestingly, we can also use this analysis to show there is no surjective group homomorphism $\mathbb{C}^* \to \mathbb{R}^*$, by using the fact that there are no nontrivial homomorphisms $\mathbb{Q} \to \mathbb{Z}/2\mathbb{Z}$ nor from $\mu(\mathbb{C}) \cong \mathbb{Q}/\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$.
EDIT: My intuition is miscalibrated: $\mu(\mathbb{C})$ is a direct summand of $\mathbb{C}^*$. This must be true because
$$ \operatorname{Ext}(\mathbb{C}^*/\mu(\mathbb{C}), \mu(\mathbb{C}))
\cong \operatorname{Ext}(\mathbb{Q}^{\mathfrak{c}}, \mathbb{Q}/\mathbb{Z}) \cong 0$$
(because $\mathbb{Q} / \mathbb{Z}$ is injective), and so the exact sequence
$$ 0 \to \mu(\mathbb{C}) \to \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \to 0$$
must be split.
Alright, The first isomorphism theorem (FIT) states that if we have an action $\phi: G\to H$ that is homomorphic, then the $ker(\phi)$ is a normal subgroup of $H$.
First, we need to show that $\phi:\mathbb R\to \mathbb T$ such that $\phi (\theta)=e^{i\pi\theta}$ where $\mathbb T:= \{z\in \mathbb C:\;||z||=1\}$. This can be easily done by letting $x,y\in \mathbb R$, So $\phi(x+y)=e^{i\pi(x+y)}=(e^{i\pi x})e^{i\pi y}=\phi(x)\phi(y)$.
Next, we need to find the kernal of $\phi$. This can be done by noting that $1$ is the identity of $\mathbb T$ so let $$\phi(\theta)=e^{i\pi\theta}=\cos(\pi\theta)+i\sin(\pi\theta)=1+0i.$$
This implies that $\cos(\pi\theta)=1$. As such $\theta = 2k$, $\forall k \in \mathbb Z$. So,
$$ker(\phi)=\{n=2k: k \in \mathbb Z\}$$
Lastly, we need to show that $ker(\phi)$ is a normal subgroup of $\mathbb T$ this is done by showing that $gH=Hg$ if $g\in ker(\phi)$. Thus,
$$\phi(n)H=e^{i\pi n}e^{i\pi\theta}=e^{i(2\pi) k}e^{i\pi\theta}=e^{i\pi\theta}=e^{i\pi\theta}e^{i(2\pi) k}=H\phi(n) $$
as desired.
Best Answer
The kernel of $f$ is the set $\{z \mid f(z) = e\}$, where $e$ is the identity of $\Bbb R_{>0}$.
Now note that $\Bbb R_{>0}$ is a multiplicative group.
Further hint: So $e$ is a number satisfying $e \cdot x = x = x \cdot e$ for all $x \in \Bbb R_{>0}$...