My understanding is that the cardinality is given by the cardinality of equivalence classes of formulae under the equivalence relation of being variants of each other i.e. identical up to uniform substitution of non-logical symbols
I get why this is at least $\aleph_0$: formulae with different numbers of non-logical symbols can't be equivalent so we need at least one equivalence class for each finite $n$
What I don't get is how, if we have $k > \aleph_0$ many non-logical symbols, there are $k$ many equivalence classes? Can someone give an example of what kinds of formulae are split into so many equivalence classes? My naive intuition is that e.g. $k$ many variable names just gives you more ways to rename a given formula in equivalent ways rather than allowing you to construct formulae which are inequivalent
I guess more concretely: can someone provide an example of two formulae which are in the same equivalence class if there are $\aleph_0$ non-logical constants, but aren't if there are $k > \aleph_0$?
Consider some language $\mathcal{L}$ of an arbitrary type $\langle (r_i)_n; (a_j)_m; \kappa\rangle$.
Let us begin with an analysis of its set of terms:
$\mathrm{TERM}_\mathcal{L} = \bigcup \mathrm{TERM}_n$, with $\mathrm{TERM}_0 = \{(c_i)_\kappa\} \cup \{(x_i)_{\aleph_0}\}$ and $\mathrm{TERM}_{n+1} = \mathrm{TERM}_n \cup \{f_j(t_1, \dots, t_{a_j}) : (t_1, \dots, t_{a_j}) \in \mathrm{TERM}_n^{a_j}, j = 1, \dots, m\}$.
Clearly, $|\mathrm{TERM}_0| = \max\{\kappa, \aleph_0\} = \mu$.
Now, if $|\mathrm{TERM}_n| = \mu$, then $|\mathrm{TERM}_{n+1}| = |\mathrm{TERM}_n| + |\mathrm{TERM}_n^{a_1}| + \dots + |\mathrm{TERM}_n^{a_m}| = \mu + \mu^{a_1} + \dots + \mu^{a_m} = \mu$. Hence, $|\mathrm{TERM}_n| = \mu$ for every $n$ and $|\mathrm{TERM}_\mathcal{L}| = \sum|\mathrm{TERM}_n| = \sum \mu = \mu$.
Now, as for its set of formulae:
$\mathrm{FORM}_\mathcal{L} = \bigcup \mathrm{FORM}_n$, with $\mathrm{FORM}_0 = \{P_i(t_1, \dots, t_{r_i}) : (t_1, \dots, t_{r_i}) \in \mathrm{TERM}_\mathcal{L}^{r_i}, i = 1, \dots, n\}$ and $\mathrm{FORM}_{n+1} = \mathrm{FORM}_n \cup \{\varphi \ \square \ \psi : (\varphi, \psi) \in \mathrm{FORM}_n^2, \square = \{\land, \lor, \rightarrow, \leftrightarrow\}\} \cup \{\neg \varphi : \varphi \in \mathrm{FORM}_n\} \cup \{Qx_i \cdot \varphi : \varphi \in \mathrm{FORM}_n, i \in \mathbb{N}, Q \in \{\forall, \exists\}\}$.
Once again,
$|\mathrm{FORM}_0| = |\mathrm{TERM}_\mathcal{L}^{r_1}| + \dots + |\mathrm{TERM}_\mathcal{L}^{r_n}| = \mu^{r_1} + \dots + \mu^{r_n} = \mu$.
$|\mathrm{FORM}_{n+1}| = |\mathrm{FORM}_n| + 4|\mathrm{FORM}_n^2| + |\mathrm{FORM}_n| + 2\aleph_0|\mathrm{FORM}_n| = \mu + 4\mu^2 + \mu + 2\aleph_0\mu = \mu$.
Hence,
$|\mathrm{FORM}_\mathcal{L}| = \sum|\mathrm{FORM}_n| = \sum \mu = \mu$.
Thus we can conclude that $|\mathcal{L}| = \mu$.
$\square$