How this equation is solved in reals

Question

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 https://mathematica.stackexchange.com/a/126156/4999 — Michael E2, Oct 25 '19 at 01:11
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 06:18

score 3 · Answer 1 · answered Oct 25 '19 at 01:40

3

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  *)

answered Oct 25 '19 at 01:40

Michael E2

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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 03:15
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 01:34

How this equation is solved in reals

1 Answers1