For example: There are two people with income less than 4K/year.
"There are at least two objects satisfying P" can be expressed in first-order logic as $$\exists x \exists y (x \neq y \wedge P(x) \wedge P(y))$$
"There are exactly two objects satisfying P" can be dealt with by first rephrasing to "There are at least two objects satisfying P, but there are not at least three objects satisfying P" and using the above idea, or alternatively as $$\exists x \exists y ((x \neq y \wedge P(x) \wedge P(y)) \wedge \forall z (P(z) \rightarrow (z=x \vee z=y)))$$
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Let's say your predicate is $P(x)$. Then here is "Exactly one $x$ satisfies $P$":
$$ \exists x. P(x)\\ \wedge (\forall y. P(y)\implies y=x)$$
That is, $P(x)$ holds, and if it holds for any $y$, then $y=x$.
Now let's do "Exactly two $x$ satisfy $P$":
$$ \exists x.\exists x'. P(x) \wedge P(x')\\ \wedge x\not= x' \\ \wedge(\forall y.P(y)\implies (y=x\vee y=x'))$$
That is, $x$ and $x'$ both satisfy $P$, they are different, and if $y$ satisfies $P$ then it must either be $x$ or $x'$.
Other formulations are possible.