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):
- $\sqrt a^{\sqrt a}=n$ for irrational $\sqrt a$
My approach: $\sqrt a^{\sqrt a}=n\Rightarrow \sqrt a^a=n^{\sqrt a}$. 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. - $a^a=n$ for algebraic and irrational $a$
This seems more likely than the first case, but still i have no idea in approaching it. - $a^a=n$ for irrational $a$
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 $a$s so that the equality will hold for natural $n$.
I would appreciate any help :)
Edit: Having Michael Hardy's answer and TomOldfield's proof for $x$ being natural OR irrational, i would still appreciate if we had something to say about $x$ being in any of these $3$ cases.
If the problem is finding a real number $x$ such that $x^x=n$, for a given $n\in \mathbb N$, then notice that the function $x\mapsto x^x$ is continuous and goes from $1$ down to a positive number less than $1$ and then up to $\infty$ as $x$ goes from $0$ to $\infty$, so the intermediate value theorem tells us a solution exists.
Finding the solution probably has to be done numerically rather than algebraically.