I was solving this question, and I'm hitting a wall.
$x^{x^4}=4$, then what is ${x^{x^8}}+{x^{x^2}}$?
Taking $\log$, $x^4\log\{x\}=\log\{4\}$, so $1<x<2$. However, I don't think that taking a $Log$ again will help me, and since there is addition in the second equation, it shouldn't work there, too. Trial and error won't help, so I have no direction in which to continue. Can anyone help?
Raising the given equation to the fourth power, we get $$(x^4)^{x^4}=4^4,$$ and that means $x^4=4,$ since $f(x)=x^x$ is monotone increasing for $x\ge e^{-1}.$ So $x^{x^8}+x^{x^2}=\sqrt{2}^8+\sqrt{2}^2=258.$
The same method works with the equation $x^{x^\alpha}=\alpha,$ but it works well only for $\alpha>1$. For $\alpha<1,$ there may be (and for $\alpha\neq e^{-1},$ there will be) a second solution, which won't be easily expressed in terms of $\alpha.$
Since $$ x^{\,x^{\,4} } = 4\quad \Rightarrow \quad \left( {x^{\,x^{\,4} } } \right)^{\,2} = x^{\,2\,x^{\,2} x^{\,2} } = \left( {\left( {\,x^{\,2} } \right)^{\,x^{\,2} } } \right)^{\,x^{\,2} } = 16 $$
then $x^2=2$ is a solution , and given that the LHS is increasing (monotone) , that is the only real positive solution.
The rest follows easily.
