Irrational to power of itself is natural I've been thinking about a natural number like n so that x^x=n for some irrational x but i couldn't find anything. As i didn't know how to approach the problem at all, i tried to make some simpler cases first (n is a natural number): 1.sqrt(a)^(sqrt(a))=n for irrational sqrt(a) My approachsqrt(a)^(sqrt(a))=n=>sqrt(a)^a =n^(sqrt(a)). And here, We suppose a is even. This means the LHS would be a natural number, and thus n^(sqrt a) is natural too. This means sqrt(a)=log_n b which i think can not be true when b is not a power of n because that time i think the logarithm would be transcendental (i'm not sure). But this only disproves the case for a being even! Still if we can prove theres no such n and a, we should check the next case. 2. a^a=n for algeb
Alexus Deleon
Answered question
2022-09-23
Irrational to power of itself is natural
I've been thinking about a natural number like so that for some irrational but i couldn't find anything. As i didn't know how to approach the problem at all, i tried to make some simpler cases first ( is a natural number):
1. for irrational
My approach: . And here, We suppose is even. This means the LHS would be a natural number, and thus is natural too. This means which i think can not be true when is not a power of because that time i think the logarithm would be transcendental (i'm not sure). But this only disproves the case for being even! Still if we can prove theres no such and , we should check the next case.
2. for algebraic and irrational
This seems more likely than the first case, but still i have no idea in approaching it.
3. for irrational
This is indeed more general than the first two cases and we should check it if the first two cases failed! Although it may be great to find all kinds of irrational s so that the equality will hold for natural
I would appreciate any help :)