TITLE: Is the logarithm of $\aleph_0$ infinite?
QUESTION [11 upvotes]: In classical mathematics $2^{\aleph_0}\ge\aleph_1$, right?
So if $2^x=\aleph_0$, what does $x$ equal?
In other words, can we define a logarithm for $\aleph_0$, and what should it be.
Is it infinite? Finite? Undefined?
Should a definition in a computational context be necessarily the same as in the classical context, or can/should it be modified, like other mathematical definitions are modified to accomodate computability (e.g. limits in the context of computable reals)

REPLY [10 votes]: There are two things to point out here.

$2^{\aleph_0}=\aleph_1$ is an assertion called "The Continuum Hypothesis". It cannot be proved, nor disproved, from the standard axioms of set theory. 
Note that in $\Bbb N$ not every number has a logarithm. For this you need to extend to the real numbers. For example $\log_25$ is not a natural number at all. So the equation $2^x=5$ has no solutions when talking about the natural numbers.
Since cardinals are not real numbers, but they do extend the natural numbers, it means that $2^x=5$ has no solutions in terms of cardinal arithmetic either. 
Why am I bringing this up? Because it shows you that you don't have to have a logarithm. And indeed for $\aleph_0$ you don't. Because if $x<\aleph_0$, then $x$ is finite and $2^x$ is finite; and if $x=\aleph_0$ then $2^x>x$ by Cantor's theorem.

As a side note, it might be worth pointing out that on a considerably large class of cardinals there is no solution to the equation $2^x=\kappa$ (where $\kappa$ is the cardinal in question).,