I've watched a lecture by Alex Wilkie on the MSRI website which covered the first order theory of exponential fields. He gave an exercise which looks very surprising. It should be solvable, however, in a few lines, and has a "very nice trick to it", as Wilkie puts it.
We consider the structure $\mathcal{C} = (\mathbb{C}, +, 0, 1, \cdot, \exp)$ with the standard interpretations and where $\exp$ is interpreted as $e^x$. The definable subsets in an algebraically closed field are well understood and as well behaved as one could possibly hope for, but the addition of the symbol for exponentiation considerably complicates things. In fact, the integers are definable by the following formula:
$\varphi(z) = \forall w (e^w = 1 \rightarrow e^{zw} = 1)$
so Gödel tells us that the definable subsets of $\mathcal{C}$ are hopelessly complicated. However, there is a conjecture of Zilber that says that this structure is quasiminimal: that is, every definable subset is either countable or cocountable.
Now, assuming this conjecture, the following remarkable fact should be true. Let $\varphi(\overline{x},y)$ be a formula. Then the set of those $\overline{a} \in \mathbb{C}^n$ such that the formula $\varphi(\overline{a}, y)$ defines a countable set is definable! I have no idea how to prove this.
I will answer myself (I hope this is okay)
Let $G$ be an (uncountable) quasiminimal group: so every subset definable with parameters is either countable or cocountable. Then a necessary and sufficient condition for a definable set $X$ to be uncountable is that $XX^{-1}=G$.
The nontrivial direction is showing that if there is some $y \notin XX^{-1}$ then $X$ must be countable. If there is such a $y$, then the translate $yX$ and $X$ are disjoint definable sets, both of the same cardinality; so by quasiminimality the set $X$ must be countable.
This finishes the exercise, as the condition $XX^{-1}=G$ is first order definable and $\mathcal{C}$ carries the structure of an abelian group under addition.