A problem given to high school students

Question

My advisor saw this problem in his son's high school textbook,

$$x^{x^{20}}=2^{\frac{1}{\sqrt{2}}}$$

How do we solve this? Note that the methods used should be something a high schooler would know.

Answer 1

Just putting form to the answer given in the comments. First, assume $x=2^a$ for some $a \in \mathbb{R}$. Then:

$$x^{x^{20}}=2^{\frac{1}{\sqrt{2}}} \implies {2^a}^{{2^a}^{20}}=2^{\frac{1}{\sqrt{2}}} $$ $$ {2^a}^{{2^a}^{20}}=2^{\frac{1}{\sqrt{2}}} \implies {{a2^a}^{20}}={\frac{1}{\sqrt{2}}}$$

Now suppose $a = 2^b$ for some $b \in \mathbb{R}$:

$${{a2^a}^{20}}={\frac{1}{\sqrt{2}}} \implies 2^b{2^{2^b20}}=2^{-\frac{1}{2}}$$ $$ 2^b{2^{2^b20}}=2^{-\frac{1}{2}} \implies 2^{2^b20+b}=2^{-\frac{1}{2}}$$ $$ 2^{2^b20+b}=2^{-\frac{1}{2}} \implies 2^b20+b=-\frac{1}{2}$$

But the solution for $2^b20+b=-\frac{1}{2}$, by guessing, is $b=-3$. Therefore: $$ b=-3 \implies a= 2^{-3}$$ $$ a=2^{-3} \implies x=2^{2^{-3}}$$ $$ x=2^{2^{-3}} \implies x=\sqrt[8]{2}$$