Let $\Bbb R^n_{++}=\{x\in\Bbb R^n\mid x_i>0, i=1,\ldots,n\}$ and let $\log\colon \Bbb R^n_{++}\to \Bbb R^n$ be the component-wise logarithmic function, i.e. $$\log(x)=(\log(x_1),\ldots,\log(x_n))\qquad \forall x \in \Bbb R^n_{++}$$ where $\log$ is the usual logarithm in base $e$ (does it make a difference?). For a subset $S\subset\Bbb R^n_{++}$, let $\log(S)=\{\log(x)\mid x\in S\}$.
For which $C\subset \Bbb R^n_{++}$, the set $\log(C)$ is convex in $\Bbb R^n$?
Note
Note that if $C$ is a line segment, i.e. $C=\{tx+(1-t)x\mid 0\leq t \leq 1\}$, then $\ln(C)$ can not be convex.
Note that for every set $C=[a_1,b_1]\times \ldots \times [a_n,b_n]$ with $0<a_i<b_i$, $\log(C)=[\log(a_1),\log(b_1]\times \ldots \times[\log(a_n),\log(b_n]$ is a convex set.
If $C=\Bbb R^n_{++}$, then $\ln(C)=\Bbb R^n$ which is convex.
You have changed the title and point of the question, by dropping the restriction that $C$ is convex. As it stands now, the answer (tautologously, and I think uninterestingly) is that $C$ must be closed under component-wise weighted geometric means, in the sense that if $x=(x_i)$ and $y=(y_i)\in C$ then so is $z=(z_i)$, where $z_i = x_i ^\alpha y_i^\beta$ for all $\alpha,\beta\in[0,1]$ with $\alpha+\beta=1$.
You might as well posit $L$ an arbitrary convex set, and then take $C$ to be the set of component-wise exponentials of elements of $C$, as in $x=(x_i)\in C$ if and only if $x_i=\exp(y_i)$ for $y\in L$.