$$\large x^{-x^{1-x}}=\large\sqrt[\sqrt2]{2}$$ $$x=?$$
Here is what I've tried:
Since RHS is $2^{\frac1{\sqrt2}}$, base of LHS should be in the form $2^a$. I checked $x=\frac12$ is an answer but not sure if there are other answers. Using $x=2^a$ left side will be equal to,
$$\large(2^a)^{\large-(2^a)^{\large(1-2^a)}}=\large(2^a)^{\large-2^{\large(a-a2^a)}}=\large2^{\large-a\times2^{\large(a-a2^a)}}$$
Hence we have $$a.2^{a(1-2^a)}={\frac{1}2}$$But don't know how to continue further.
Let $y=x^{-x}$. Then $$x^{-x^{1-x}}=y^y=\sqrt[\sqrt{2}]{2}=\sqrt{2}^\sqrt{2}>1$$ Now the equation $$z^z=w$$ has one solution if $w\geq 1$, and two if $e^{-1/e}<w<1$. Hence, $$y=\sqrt{2}$$ Rewriting in terms of $x$, $$x^x=(1/2)^{1/2}<1$$ It follows that $x=1/2$ is one solution, and it is easy to check that $x=1/4$ must be the second solution.