Is $\log(\aleph_0)$ undefined?

Question

NOTE: In the context of this question, the base of $\log(x)$ is $10$.

I was researching cardinal arithmetic when I found out about logarithms of infinite cardinal numbers. Assuming the axiom of choice and given a infinite cardinal $κ$ and a finite cardinal $μ$ greater than 1, there may or may not exist a cardinal $λ$ which satisfies $μ^κ=λ$. If $λ$ does exist, then it is infinite and less than $κ$.

Since $\aleph_0$ is the smallest infinite cardinal number, $λ$ cannot exist, so does that mean $\log(\aleph_0)$ is undefined?

Answer 1

I don't think $\log \aleph_0$ exists.

An easy way to see this is that if we denote the power set of $\mathbb A$ as $2^\mathbb A$, then there is no cardinal $\mathbb X$ such that $\aleph_0=2^\mathbb X$ because $\aleph_0$ is the smallest cardinal by definition. So, there is no cardinal $\mathbb X$ such that $\log \aleph_0=\mathbb X$.

Answer 2

I think it doesn't exist because,

Let, $X=\operatorname{log}(\aleph_0)$

$\aleph_0=10^X$

If, $X\lt\aleph_0$

Then, $10^X\lt\aleph_0$

So, $X\not\lt\aleph_0$

And if, $X\geq\aleph_0$

Then, $10^X\geq 10^{\aleph_0}=2^{\aleph_0}\geq\aleph_1\gt\aleph_0$

So, $X\not\geq\aleph_0$

This leads us to the only conclusion that, $\textstyle\displaystyle{X\in\emptyset}$.

So, $\operatorname{log}(\aleph_0)$ cannot exist.