# Thread: infinite exponentials

1. ## infinite exponentials

Hi all,
A friend asked me if i could sove Ln(x)=-x in exact form.
The solution i gave was x=e^(-e)^(-e)^(-e)^...
Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
A similar problem is x^x^x^x....=2 (or 4)
Which can be solved as x^2=2 x=root 2
or x^4=4 x=root 2 also

I would apreciate any thoughts on this and infinite exponention. All proofs are welcome.

2. Originally Posted by AndrewAllen
Hi all,
A friend asked me if i could sove Ln(x)=-x in exact form.
The solution i gave was x=e^(-e)^(-e)^(-e)^...
Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
First, note that "a^b^c" must be interpreted as a^(b^c) because (a^b)^c= a^(bc) which gives nothing new. If x=e^(-e)^(-e)^(-e)^... then, taking e to the negative power of each side, e^-x= e^-(e^(-e)^(-e)^(-e)^...)= x (since the right side is an infinite tower of "-e"s still). Solve the equation e^-x= x, which is, of course, the same as -x= ln(x). There is no algebraic way to solve such an equation but numerical methods give x as approximately 0.55

A similar problem is x^x^x^x....=2 (or 4)
Which can be solved as x^2=2 x=root 2
or x^4=4 x=root 2 also

I would apreciate any thoughts on this and infinite exponention. All proofs are welcome.
For the second, if there were an x satisfying x^x^x^x...= 2, then taking x to the power of each side, x^(x^x^x^x...)= x^2 or x= x^2. No, that is NOT "x^2= 2 x= root 2". It is x^2- x= x(x- 1)= 0. That has roots 0 and 1. But 0^0 is not defined and 1^(1^(...)) is 1 no matter how many powers you take, certainly not 2. There is no x satisfying that equation.

3. Originally Posted by HallsofIvy

For the second, if there were an x satisfying x^x^x^x...= 2, then taking x to the power of each side, x^(x^x^x^x...) = x^2 or x= x^2. No, that is NOT "x^2= 2 x= root 2". It is x^2- x= x(x- 1)= 0. That has roots 0 and 1. But 0^0 is not defined and 1^(1^(...)) is 1 no matter how many powers you take, certainly not 2. There is no x satisfying that equation.
But isn't the red just 2 so x^2 = 2?

This problem is a famous problem dating back to Euler 1778. Here's a website with more information

Hyperpower Function

4. Oops!

5. Originally Posted by AndrewAllen
Hi all,
A friend asked me if i could sove Ln(x)=-x in exact form.
The solution i gave was x=e^(-e)^(-e)^(-e)^...
Ln(x)=Ln[e^(-e)^(-e)^...]=(-e)^(-e)^(-e)^...=-x

Whilst exploring this answer i have had difficulty believing this to be well defined or even correct.
[snip]
$\displaystyle \ln x = -x \Rightarrow x = e^{-x} \Rightarrow x e^x = 1 \Rightarrow x = W(1)$ where W is the famous Lambert W-function:

http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf

Lambert W function - Wikipedia, the free encyclopedia

Lambert W-Function -- from Wolfram MathWorld