# Proof 2^x = x

If 2^x = x

then

2^(2^x) = x

also!

2^2^(2^x) = x

2^2^2^(2^x) = x

2^2^2^2^2^2^2^ ...^(2^x)_n = x

As russ wrote to Dr. Math
On 04/03/2004 at 05:43:34 (Eastern Time),
>[Question]
>The equation 2^x = x .
>
>2^x = x
>
>then
>
>2^(2^x) = x
>
>2^2^(2^x) = x
>
>2^2^2^(2^x) = x
>
>2^2^2^2^2^2^2^2^ ... 2^(2^x)_n
>
>How can this equation be solved symbolically?
>
>
>[Difficulty]
>
>
>[Thoughts]
>2^x = x
>
>2 = x^[1/x]
>
>x^[1/x]- 1 - 1 = [2^x]/x ...

Dear Russ,
What is the scope of your investigation? Are you
looking for
answers or methodology? Are you interested in complex
solutions or
only real numbers?
to the power x to
the power x are astute, and could be one way to look
for a solution.
There are other methods as well, but none that
involves only a finite
number of calculations. The only solutions to this
kind of equation
are as a limit to an infinite process.
You may be interested in this chapter from our
archives:
http://www.mathforum.org/library/drmath/view/53229.html

NateTG
Homework Helper
Hmm.
$$\frac{d}{dx} 2^x = \ln 2 2^x$$
which is monotone increasing and
$$\frac{d}{dx} x = 1$$
which is constant, so if
$$2^x > x$$
where
$$\ln 2 \times 2^x = 1 \rightarrow x= -\log_2 ({\ln 2}) \approx 0.5$$
then the only solutions are imaginary.
So I don't think there are any real solutions.

Huh?

If 2^x=x, then

2^(2^x) does not equal x, it equals 2^x; you have to do the same thing to both sides, right??? I guess I don't understand your question then. It definitely does not have real solutions, because of the reason above.

Hurkyl
Staff Emeritus