Question

y=x^x^x^x... where x=sqrt(2), solve for y

Answer 1

$$
\Large{
y=x^{x^{x^{x^{x^{x^\cdots}}}}}\\
\text{where } x=\sqrt2\\
\text{Solve for y}}
$$

Answer 2

Hint:
$$
\Large{
x^y=y
}
$$

Answer 3

\(\sf\Large (\sqrt{2})^2=2\)

So, therefore, y must be equal to 2.

Sorry, I'm giving a direct answer, but considering from your hint, this is a challenge problem for others. :)

Answer 4

but \[\sqrt{2}^4=4\] ;)

Answer 5

Hint2: Take logs on both sides
Put in form
$$
W(y)\exp^{W(y)}
$$

Answer 6

y = x^y => logy = y log x
=> log y ^1/y = log 2^1/2
=> y =2

Answer 7

2 is the right answer, but \(\log y^{1/y}=y^{1/y}\log x\) and at \(x=\sqrt 2\) \(\log y^{1/y}=y^{1/y}\log \sqrt{2}\) how does y=2 follow?

Here's what I was thinking:
$$
x^y=y\\
y\ln x=\ln y\\
\ln x=\cfrac{\ln y}{y}=\cfrac{\ln y}{\exp^{\ln y}}=\ln y \exp^{-\ln y}\\
-\ln y=-\ln x \exp^{-\ln x}=W(x)\exp^{W(x)}\\
-\ln y=W(-\ln x)\\
y=\exp^{-W(-\ln x)}
$$
Using the definition of \(W()\)
$$
W(x)\exp^{W(x)}=x\\
\exp^{W(x)}=\cfrac{x}{W(x)}
$$
and
$$
\exp^{nW(x)}=\left(\cfrac{x}{W(x)}\right)^n\\
$$
$$
y=\exp^{-W(-\ln x)}=\left(\cfrac{-\ln x}{W(-\ln x)}\right )^{-1}=-\cfrac{W(-\ln x)}{\ln x}
$$
For \(x=\sqrt 2\)
$$
y=-\cfrac{W(-\ln \sqrt 2)}{\ln \sqrt 2}=2
$$
Where W() is the Lambert W-function http://dlmf.nist.gov/4.13

http://www.wolframalpha.com/input/?i=-productlog%28-ln%28sqrt%282%29%29%29%2F%28ln+sqrt%282%29%29