Solving the Infinite Exponential Equation: Does x Exist?

[Kittel Knight]

Consider the equation

$$x^{x^{x^{...}}} = 2$$

Does x exist ?

Well, at first, I would say x=sqrt(2) , but is this ok?

In general, x^x^x^... = Z would imply x = Z^(1/Z)
But, if x>1, then x^x^x^... is crescent.
In other words, when "Z" increases, then "x" increases.

However, lim oo Z^(1/Z) = 1
So, if x^x^x^... increases, it means that "x" goes to 1 ?!

Where is the mistake?

 

[futurebird]

x must be between $$e^{-e}$$ and $$e^{\frac{1}{e}}$$. Euler gave a proof showing that it only has a limit in this interval.

I don't know if the answer is right, but it's in the interval... so it's possible.

 

[real10]

I reduced it to
x^k * ln(x) = ln(2) for some k
so i guess find x and k? like x= 2 and k = 0...
dunno if this right just wondering...

 

[morphism]

It can be reduced to x^2 = 2 immediately.

 

[real10]

It can be reduced to x^2 = 2 immediately.

could u explain for some reason I am not able to see it?
Thanks!

 

[Dragonfall]

$$x^{x^{x...}}}=x^2=2$$

 

[HallsofIvy]

In case you are wondering, Dragonfall is taking x to the power of each side.
x to the $x^{x^{x...}}}$ is equal to $x^{x^{x...}}}$ since there were an infinite number of "x"s to begin with and according to the equation, that is equal to 2. On the right side, of course, x to the 2 power is x²: 2= x².

 

[real10]

In case you are wondering, Dragonfall is taking x to the power of each side.
x to the $x^{x^{x...}}}$ is equal to $x^{x^{x...}}}$ since there were an infinite number of "x"s to begin with and according to the equation, that is equal to 2. On the right side, of course, x to the 2 power is x²: 2= x².

makes sense overlooked the fact that there are infinite number of them..
thanks.