What is the solution for x in the infinite exponent tower x^x^x^...=2?

[saltydog]

Thanks Daniel,
However, my understanding is that the hyperexponent function above converges only when x is in some interval. Can you please tell me how to incorporate math symbols into my posting so I can be more specific? I have Mathematica. Can I export data from there?

Thanks,
SD

Ok Daniel,
I checked out the FAQ and found LaTeX. Nice feature here!
SD

 

[Physics is Phun]

I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could someone post a sollution?

 

[StatusX]

I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could someone post a sollution?

If you mean how do you find that the answer is sqrt(2), look at my post above. If you mean how could it be that a number greater than 1 converges in an infinite tower, which at least I found strange, just think of it like this: Sqrt(2)^2 is 2, so sqrt(2)^x for any x less than 2 will be less than 2. This means that sqrt(2)^sqrt(2)<2, and so then sqrt(2)^(sqrt(2)^sqrt(2))<2, and so on.

 

[The Divine Zephyr]

Though I am still alittle quesitonable about sqrt 2 as an answer, I think I understand the reason the answer is correct.