Is the logarithm of $\aleph_0$ infinite?
A simple answer would be that if we interpret $2^A$ as denoting the powerset of $A$, there should be no $X$ such that $\mid2^X\mid = \aleph_0$. The cardinality of $X$ itself cannot be $\aleph_0$ (or we would have $\mid2^X\mid > \aleph_0$), but it also cannot be any finite number, or we would have $\mid2^X\mid$ also finite.
In the absence of an infinite cardinal smaller than $\aleph_0$, a set with the cardinality of $X$, as defined here, cannot exist.
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.
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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).,
The logarithm of an infinite cardinal number $\alpha$ is defined as follows: $$\log\alpha=\min\{\beta\ |\ 2^\beta\ge\alpha\}.$$ This definition is given on p. 74 of Cardinal Functions in Topology by István Juhász, and I am not aware of any competing definitions. According to this definition, $$\log\aleph_0=\log\aleph_1=\aleph_0.$$ It should be noted that $$2^{\log\aleph_0}\gt\aleph_0.$$ The equation $2^x=\aleph_0$ has no solution.