Can I find $x^{x^{-1}}$ given $x^{-17^{ 17^{-x}}}=17$?

Question

Has This problem a solution? I tried for many ways, even Mathematica. I think it has a typo:

$$x^{-17^{ 17^{-x}}}=17$$

Find: $$x^{x^{-1}}$$

How can I explain it has no solution.

Kind regards.

Answer 1

Let $a\in \mathbb R$ be a solution to the equation $t=17^{-t}$. It is easy to prove that such a solution exists and is unique. Indeed, $f(t):=t-17^{-t}$ is continuous, $f(0)<0$, and $f(1)>0$, so there must exist $a\in[0,1]$ such that $f(a)=0$; the uniqueness follows from the fact that $f$ is a strictly increasing function (since $f'(t)=1+17^{-t}>0$ for all $t\in \mathbb R$).

Now let us prove that $x=a$ is also a solution to the more complicated equation $$ g(x):=x^{{-17}^{17^{-x}}}=17. \tag{1} $$ Indeed, $$ 17^{-a}=a \Rightarrow g(a)=a^{-17^{a}}, \tag{2} $$ $$ 17^a=a^{-1}\Rightarrow g(a)=a^{-a^{-1}}, \tag{3} $$ $$ a^{-1}=17^a \Rightarrow (a^{-1})^{a^{-1}}=(17^a)^{a^{-1}}\Rightarrow g(a)=17. \tag{4} $$ Finally, from $(3)$ and $(4)$ it follows that $$ a^{a^{-1}}=\frac{1}{g(a)}=\frac{1}{17}. \tag{5} $$