$\forall A\exists L(\mathcal P(L)=A)$ To be honest this idea is not mine but i saw this axiom somewhere and I dont remember where
Is the axiom that say that exist the "logarithm set" $L$ for every set $A$.

"$\forall A\exists L(\mathcal P(L)=A)$"

Using the intuition in tha naive set theory we can see that such $L$ ca not exist always and we can derive some contradiction too (in the naive concept) like for example this
we have that
$\forall X (\varnothing \subset X)$ then $\forall X (\varnothing \in \mathcal P(X))$
But if we allow the logarithm set axiom we have
$\forall A\exists L(\mathcal P(L)=A)$ imply $\forall A(\varnothing \in A)$
and is not true.
In addition we have that the logarithm set $L$ of $A$ always belongs to $A$.
$\forall A (\varnothing \in A)$
If I understand well this means that the logarithm aximom is useless beacuse states the existence of an element $L$ of the set $A$ for every set $A$, then is in contraddiction with the empty set axiom.
At this point I must ask why and when this axiom can be used:

1- Do the contraddictions that I've found are enough for "kicking out
of the game" this axiom for ever?
2-There can be an alternative
collection of set theory axiom that can allow this axiom to be
interesting? Maybe we can extend the universe of the set with "exotic"
sets that make us always able to  take the "Logarith set" of every
set?

Sorry if I made some grammar errors and thanks in advance.
 A: Yes, the contradiction that you have found is sufficient to conclude that this axiom is very contradictory (i.e. it is a contradiction to one of the fundamental properties of set theory, namely the empty set).
One can perhaps modify this to be an argument of cardinality. That is, $$\forall A\exists L(A\neq\varnothing\rightarrow|2^L|=|A|),$$ but that is also not going to work. Note that the integers are not closed under taking $\log_2$. If you want this axiom to make sense somehow then you need to circumvent this fact. There will never be any set whose power set contains exactly three elements.
So maybe if we require that $A$ is infinite? But that also going to fail. $\aleph_0$ has the property that $|2^X|\neq\aleph_0$, no matter what $X$ is. And it is not the only infinite cardinal that has this property. So you're going to try and work harder and harder on that. This will also negate the existence of inaccessible cardinals, which have the same property as $\aleph_0$ here.
So let's assume that we want to maximize the class of sets equipotent with power sets. What do I mean by maximize? Well given an infinite cardinal $\aleph_\alpha$, I want that the cardinals smaller than $\aleph_\alpha$ which are equipotent with power sets is as large as possible without causing contradictions for smaller $\alpha$'s. Assume that we can formalize this, let us observe the consequence and then see what is the correct formalization.
So we know what happens with finite sets, and we know that not all infinite sets can be equipotent with power sets. $\aleph_0$ can't be. But $\aleph_1$ can be. In that case $2^{\aleph_0}=\aleph_1$. Then we want $\aleph_2$ to be a power set, and so $2^{\aleph_1}=\aleph_2$. We can continue. It is not hard to see that by the time we reached $\aleph_\omega$, the first limit cardinal, $\sf GCH$ holds below it. We have to skip $\aleph_\omega$ because it provably can't be equipotent to a power set. So we continue in a similar fashion, and it is apparent that by the time we reached $\aleph_{\omega_1}$ that the power sets of sets of smaller cardinality are exactly $\aleph_{\alpha+1}$ for $\alpha<\omega_1$.
While $\aleph_{\omega_1}$ is eligible as a power set (e.g. it is consistent that $2^{\aleph_0}$ is $\aleph_{\omega_1}$), since we already have that it is a strong limit cardinal -- it can't a power set. We can continue on and on, so we have that in fact $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ for every ordinal $\alpha$.
Essentially, we require $\sf GCH$ to hold. That we can formalize.
