Here, $\langle a \rangle$ denotes the smallest subgroup of $\mathbb{Q}_{>0}$ containing $a$.
I've already proved this by showing that $\langle 2 \rangle$ and $\langle 3 \rangle$ are both isomorphic to $\mathbb{Z}$, but I'd like to prove it by showing that a well defined function $f: \langle 2 \rangle \to \langle 3 \rangle$ is bijective and operation-preserving. Since $\langle 2 \rangle = \{2^n:n\in\mathbb{Z}\}$ and $\langle 3 \rangle = \{3^n:n\in\mathbb{Z}\}$, I defined $f$ such that $f(2^n) = 3^n$. How do I now show that $f$ is well defined by writing $f(x) = \dots$? I know it probably includes a log and I can perhaps guess what the explicit function might be, but I was wondering if there was a process like 'solving for x' but in the context of writing an explicit function.
You don't need a complicated expression but if you want one, here it is: $$ f(x) = 3^{\log_2(x)} $$
However, this expression only makes sense in $\langle 2 \rangle$, because for the other rationals, the value of $f$ will be irrational.
Bottom line: stick to the simple $f(2^n)=3^n$.