In the real numbers, we always have that $\log(\exp(x)) = x$ for all $x\in\mathbb{R}$. However, as How to prove $\log(\exp(z))\neq z$ if $z\in \mathbb{C}$ shows, we can find some $z\in\mathbb{C}$ such that $\log(\exp(z))\neq z$ depending on the branch of $\log$ that we use. So if I have an equation $f(z) = \exp(g(z))$, and I want to isolate $g(z)$, I would just take logarithms and be done if I were working in $\mathbb{R}$. But if I am working in $\mathbb{C}$, I can't be assured that $\log(\exp(g(z)) = g(z)$, so how could I accomplish isolating $g(z)$? I assume that I would have to choose a specific branch cut to accomplish this but am not certain which one to choose.
NOTE: Assume that I have no prior knowledge of what values $f(z)$ takes on.
$$\log f(z) = \log |f(z)| + i \arg f(z)$$
$$\log |f(z)| = \log \exp( \Re(g(z))) = \Re(g(z)).$$
$$\arg f(z) = \arg \exp(g(z)) = \arg \exp(i \Im g(z)).$$
Yes, you'll need to define your branch to compute the argument, and then
$$ \log f(z) = \Re(g(z)) + i \arg \exp(i \Im g(z)).$$
$$ \log f(z) = \Re(g(z)) + i (\Im g(z)+ 2k\pi )\quad (k \text{ depends on the branch}).$$
For the example you allude to:
$$g(z) = z = x+ i y,$$ $$\log f(z) = x + i \arg \exp( i y) = x + i (y + 2k\pi), \quad (k \text{ depends on the branch}).$$