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ybarrap

  • one year ago

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

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  1. ybarrap
    • one year ago
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    $$ \Large{ y=x^{x^{x^{x^{x^{x^\cdots}}}}}\\ \text{where } x=\sqrt2\\ \text{Solve for y}} $$

  2. ybarrap
    • one year ago
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    Hint: $$ \Large{ x^y=y } $$

  3. AbdullahM
    • one year ago
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    \(\sf\Large (\sqrt{2})^2=2\) 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
    • one year ago
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    but \[\sqrt{2}^4=4\] ;)

  5. ybarrap
    • one year ago
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    Hint2: Take logs on both sides Put in form $$ W(y)\exp^{W(y)} $$

  6. princeharryyy
    • one year ago
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    y = x^y => logy = y log x => log y ^1/y = log 2^1/2 => y =2

  7. ybarrap
    • one year ago
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    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

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