What is solution to
x^(x^(x^(x^x^...)))....)))) = 2?
( That is x raised to x raised to x ..infinite times... = 2)

One immediate thing comes to mind is:
it is same as x^(the whole thing which is 2) =2
so x = sqrt 2

but if we consider,
x^(x^(x^(x^x^...)))....)))) = 4
still we get (if apply same logic)
x=sqrt 2
but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?
Surely if x=sqrt2 this series should blow up..and can't remain finite.....
Then what is the way to find x?

I think there is a result due to Euler that the equation
x^(x^(x^(x^x^...)))....)))) = c has a real solution iff 1/e <= c <= e.

If x = -1, x^(x^(x^(x^x^...)))....)))) has already converged.   ;-)

Here is a hint:

z = 1000 does not work,
nor z = 1000000, nor...

One cannot evaluate this -
  x^(x^(...)) = 2
because there is no initial value to start with the
evaluation.

I think you (I'm adressing the OP here) should write
  ...^(x^(x)) = 2
and then
  ...^(x^(x)) = y
  x^y = y

Now you may search for x such that x,x^x,x^x^x,...^x^x^x
converges and find a range for x, where this converges.
(Euler-range for x):

 1/e^e < x < e^(1/e)
or
 -1*e    < ln(x) <  e^-1

Insert this and you find
 x = 2^(1/2) = sqrt(2)
which is in the (Euler-) range.

Now you say, but the same x can be used to express:

  ...^(x^(x)) = 4
because
    x^4 = 4
    x = 4^(1/4)

But how would you introduce a function (of x), which provides
two values for the same x? Functions are *defined* to provide
exactly one value for each x (if at all) .
And also: you may try any value for x from the convergent
Euler-range and approximate- you'll never arrive at 4.

It's the same as with x^2 = y

Formulated as a *function* x=f(y)
 x = f(y)=  y^(1/2)
you have only one solution x for each y (for y=2 we have x = 1.414... )
As a function in x
 y = f(x) = x^2
we have then one value for y for each x (where different x may give
the same value for y, for x0=-sqrt(2), x1=+sqrt(2) y0=y1=2 )

Using the above problem it is

 x = f(y) = y^(1/y)

By the definition of a function we have one solution x for each y
(which may be arrived by different y, namely x=sqrt(2) for y=2 and y=4).

If we want to understand this as *function* of x
 y =  f(x) = lim ...^(x^(x^x))
then by definition we can have only one value of y for any x by the
definition of a function.
(Again: many x may give the same value for y - but this was not
asked in the OP)

If we want to understand y=...^(x^(x^x)) as *inverse* f^-1(y) of
 x = f(y) = y^(1/y), with a *set of solutions* in y we have to
introduce a new convention, different to that of functions.

For the inverse of the exponential-function,  y = f(x) = exp(x),
we have the same problem, we find not only two (as in the
sqrt-example) but even infinitely many (complex) values x giving
the same y:
we have a set of values of x mapping to the same y : {x0,x1,x2,...} -> y
and a notion for an inverse of this mapping y-> {x0,x1,x2,...} needs
then the introduction of another concept; often simply formulated as
"multivalued function".
I find this an unlucky formulation, and think, there should
be a better name for this. Maybe this name was chosen, because
for y=exp(x) the set of {x0,x1,x2,...} has a simple structure:
xk = x0 + k*2*pi*I  , k=0..inf
(But that's only speculation, don't really know the exact whereabouts
of the introduction of the name "multivalued function" besides Euler's
treatise of the log-function, where he dealt with this)

---
Hmm, my own argumentation still doesn't look very straight, as I read
this second time before posting, sorry; I seem to need some more practice
with this discussion myself...

Gottfried

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Regards,
Sujit P Gujar.
IISc Bangalore.
Web : http://people.csa.iisc.ernet.in/sujit