This is the following problem: Let $X$ be a set. Prove that there is not a surjection from $X \rightarrow 2^X$ (Hint: Assume to the contrary that $f\colon X \rightarrow 2^X$ is a surjection and consider the set $M=\{ x \in X | x \notin f(x) \}$). Use these facts to conclude that there is an infinite number of sizes of infinity cardinals.
My answer to the question is: This function is only bijective if its inverse is bijective and since $f(x)=\log(x)$ isn't bijective then the original function $f(x)=2^x$ isn't bijective. Since this function isn't bijective, then it isn't surjective.
As for the cardinality, I'm not sure what that really means. Thank you in advance for any hints or help!
Suppose there is a surjection $f :X \to 2^X$. Recall that $2^X$ is the power set of $X$, and so consists of subsets of $X$.
As the hint suggests, consider the subset $M \subseteq X$ such that $M = \{x \in X : x \not \in f(x) \}$. Since $M \subseteq X$, $M$ is an element of $2^X$, and since $f$ is meant to be surjective, this means that there should be some $y \in X$ such that $f(y) = M$.
This gives us a contradiction however, because either $y \in M$, or $y \not \in M$. In the first case, the defining property of $M$ says that $y$ cannot be in $M$, giving a contradiction. In the second case, since $y \not \in M$, it must satisfy $ y \in f(y)$, again by the defining property of $M$. But this gives a contradiction since we've already said that $f(y) = M$.
Thus it follows that there can be no surjective map from $X$ to $2^X$.
Applying this argument to the case $X = \mathbb{N}$, we get a set $X_1$ with strictly larger cardinality than $\mathbb{N}$. Applying the argument to $X_1$ gives you another set $X_2$, with strictly larger cardinality than $X_1$. A simple induction argument will then give you an infinite family $\{X_n\}$ of infinity cardinals.