Well, Yea Forums?

But that's wrong.

Enlighten us

X^2=2 hence X=sqrt2.

When you have a pattern almost always the solution is the mathematical inversor.

I'm not that homeschooled user. Just answering for him. Anyone can be a teacher, not everyone can be an effective one.

sqrt(2) ^ sqrt(2) ^ sqrt(2) ^ sqrt(2) is already greater than 2
sqrt2 ^ sqrt2 ^sqrt2 ^ ... goes to infinity

Don't really have time to show the derivation, but the most approximate answer yields x = 0.5ln(exp(Σ(1/2^n)))

Well yeah I agree

Clearly the ... leads off to a X^X+1
You can't really expect me to believe it goes on forever.

That fancy parentheses is useless. You exponentiate then take the logarithm so they cancel each other out. The sum of inverse powers of 1/2 is 2 or 1 depending on whether n=0 or 1 as starting value. Half those is 1 or 1/2. Neither are solutions

>sqrt(2) ^ sqrt(2) ^ sqrt(2) ^ sqrt(2) is already greater than 2
How sure are you about that?