Question.
Suppose $\displaystyle x^{-81^{81^{-x}}}=81$. Find $\sqrt[4x]{x}$.
Attempt.
As far as I know, when we face this kind of problems in preparation for the university entrance exam in Latin America, we should take powers to a convenient value and try to resemble the exponents by their factors to look the same as the term in left. For example, converting $$(ax)^x = a \Rightarrow (ax)^{ax}=a^a$$ Let us solve for $ax=a$, then $x=a^{a-1}$. But I can't do that here. I tried making it $A^A$, where $A=81^{-x}$, but the roles of exponent and base are interchanged in the problem, so it doesn't work.
We have
$$81^{-1} = x^{81^{81^{-x}}},$$
and hence
$${\left(81^{-1}\right)}^{81^{a}} = {\left(x^{\left(81^{81^{-x}}\right)}\right)}^{81^a} = x^{\left(81^{{81^{-x}}}\right)\cdot \left(81^a\right)} = x^{\displaystyle 81^{\left({81^{-x}}+a\right)}}.$$
Therefore, with $a = -81^{-x}$ we have
$${\left(81^{-1}\right)}^{81^{-81^{-x}}} = x^{81^0} = x,$$
or
$$x = 81^{-81^{-81^{-x}}}.\tag{$*$}$$
Substituting $x$ repeatedly in the expression above would gives us
$$x = 81^{-81^{-81^{-81^{\dots}}}}, \tag{1}$$
and if this made sense we could identify $x = 81^{-x}$. It would then follow that $\sqrt[4x]{x} = \frac13$.
We can be more rigorous as follows. Let $f:\Bbb R\longrightarrow \Bbb R$ be given by $f(t) = -(81^t)$. Observe that $t = 81^{-t} \iff -t = -\left(81^{-t}\right) \iff -t$ is a fixed point of $f$.
Now, $f$ is a strictly decreasing function, and it's thus easy to see that it has a single fixed point. We can rewrite equation $(*)$ as follows:
$$-x = -\left(81^{-\left(81^{-\left(81^{-x}\right)}\right)}\right) = f^3(-x),\tag{$**$}$$
so $-x$ is a fixed point of $f^3$. Of course, the fixed point of $f$ is also a fixed point for $f^3$.
It remains to check that $f^3$ also has a single fixed point, from which it follows that $-x$ must be the fixed point of $f$. This once again follows from the fact that $f^3$ is strictly decreasing; if this is not clear, it's a good exercise to look at how monotonicity behaves with compositions of strictly decreasing and strictly increasing functions.
We've shown hence that $x = 81^{-x}$, so that indeed $\sqrt[4x]{x} = \frac13$ as we had guessed before.
For what it's worth, $x = 81^{-x}$ has a solution in terms of Lambert's $W$, given by
$$x = \frac{W(\log(81))}{\log(81)}\approx 0.285360471863842099782478678983274970610868803768259$$
One can check that the approximation above 'satisfies' the equation
$$x^{-81^{81^{-x}}}=81$$
of the opening statement.