ybarrap one year ago y=x^x^x^x... where x=sqrt(2), solve for y

1. ybarrap

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

2. ybarrap

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

3. AbdullahM

$$\sf\Large (\sqrt{2})^2=2$$ 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. :)

4. Zarkon

but $\sqrt{2}^4=4$ ;)

5. ybarrap

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

6. princeharryyy

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

7. ybarrap

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