# How this equation is solved in reals

I have tried to solve this equation $$2^{-2 ^{2^{-x}}}=2$$ using

FindInstance[2^(-2^(2^(-x))) == 2, x, Reals]


The result is

{{x -> Root[{-2 Log[2]^2 + 2^(1 + #1) Log[2]^2 #1 &, 0.64118574450498598449}]}}


then find in value of x^(1/x)

I understand that the result is 1/2 but I can't prove it.

What does the # in the result mean?

• Look up Root in the docs and also see mathematica.stackexchange.com/a/126156/4999 – Michael E2 Oct 25 '19 at 1:11
• No real solution exists in the range -1<x<1 : NSolve[{2^(-2)^2^(-x) == 2, -1 < x < 1 }, x, Reals] (*{}*) – Ulrich Neumann Oct 25 '19 at 6:18

For positive real x, x^(1/x) == 1/2 is equivalent to x == 1/2^x:

x == 1/2^x /.
{x -> Root[
{-2 Log[2]^2 + 2^(1 + #1) Log[2]^2 #1 &,
0.64118574450498598449}
]} // FullSimplify

(*  True  *)


Or you can get numerical evidence:

N[
x^(1/x) /.
{x -> Root[
{-2 Log[2]^2 + 2^(1 + #1) Log[2]^2 #1 &,
0.64118574450498598449}]},
10]  (* number of digits of precision to compute *)

(*  0.5000000000  *)

• E^(-ProductLog[Log[2]]) == ProductLog[Log[2]]/Log[2] == Root[{-2 Log[2]^2 + 2^(1 + #1) Log[2]^2 #1 &, 0.64118574450498598449}] // FullSimplify evaluates to True – Bob Hanlon Oct 25 '19 at 3:15
• What does the # in the result mean? – zeros Oct 25 '19 at 23:47
• @zeros - # (Slot) is an argument in a pure function. Read the documentation for Function – Bob Hanlon Oct 26 '19 at 1:34