So for instance, if you take the Beth numbers as a sequence of infinite cardinal numbers:
$\beth_0=\aleph_0$,
$\beth_1=\mathcal{P}(\beth_0)$,
$\beth_2=\mathcal{P}(\beth_1)$,
etc., where $\beth_{\alpha+1}=2^{\beth_\alpha}$ (https://en.wikipedia.org/wiki/Beth_number). I was wondering if you can "undo" a powerset by taking the $\log_2$ of the set. Therefore, we could see that, for example, $\log_2(\beth_1)=\beth_0$, since $\beth_1=2^{\beth_0}$. (Since $\log_2x$ "undoes" $2^x$). However, if we try to apply this to $\beth_0$ (which is equal to $\aleph_0$), by taking $\log_2(\beth_0)$ we should get some cardinal number $\beta$, such that $2^\beta=\beth_0$.
However, this poses a seeming paradox: if $\beta$ is finite, then that means we can take $2$ to the power of some finite number to get an infinite number, which we know cannot be true. Therefore, $\beta$ has to be infinite–but what infinite cardinal is it? It can't be larger than $\beth_0$, so it must be $\beth_0$? But this implies that $\log_2(\beth_0)=\beth_0$, and this further implies that $2^{\beth_0}=\beth_0$. But we know that by definition, $2^{\beth_0}=\beth_1$. So, how does this work? I'm stuck and I'm sure that I'm just overlooking something simple but I can't figure out what it is.
Best Answer
The boring answer is just that it doesn't make sense to talk about the "$\log_2$" of an infinite cardinal.
One problem is the one you discovered: some cardinals, like $\aleph_0$, are not equal to $2^\kappa$ for any cardinal $\kappa$. So there's no obvious way to define the logarithm of such a cardinal. To put it another way, the exponential function $\kappa\mapsto 2^\kappa$ is not surjective on infinite cardinals.
Even worse, it's consistent with ZFC that the exponential function fails to be injective on infinite cardinals. For example, we could have $2^{\aleph_0}=2^{\aleph_1}$. What should $\log_2$ of this cardinal be?