A community for students.
Here's the question you clicked on:
 0 viewing
ybarrap
 one year ago
y=x^x^x^x... where x=sqrt(2), solve for y
ybarrap
 one year ago
y=x^x^x^x... where x=sqrt(2), solve for y

This Question is Closed

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0$$ \Large{ y=x^{x^{x^{x^{x^{x^\cdots}}}}}\\ \text{where } x=\sqrt2\\ \text{Solve for y}} $$

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Hint: $$ \Large{ x^y=y } $$

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.0\(\sf\Large (\sqrt{2})^2=2\) 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. :)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1but \[\sqrt{2}^4=4\] ;)

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0Hint2: Take logs on both sides Put in form $$ W(y)\exp^{W(y)} $$

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0y = x^y => logy = y log x => log y ^1/y = log 2^1/2 => y =2

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.02 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 Wfunction http://dlmf.nist.gov/4.13 http://www.wolframalpha.com/input/?i=productlog%28ln%28sqrt%282%29%29%29%2F%28ln+sqrt%282%29%29
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.