Is 2^x = x the Key to Infinite Exponential Solutions?

[PFanalog57]

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?
Your thoughts about an infinite series, raising x
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]

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.

 

[philosophking]

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]

If 2^x=x, then

2^(2^x) does not equal x, it equals 2^x;

Sure, 2^(2^x) equals 2^x... but what does 2^x equal?

 

[philosophking]

Ah i see, defined recursively

 

[phoenixthoth]

x=-LambertW(-log2)/log2, i think...