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Infinite exponential

  1. Oct 11, 2007 #1
    Consider the equation

    [tex]x^{x^{x^{...}}} = 2 [/tex]

    Does x exist ?

    Well, at first, I would say x=sqrt(2) , but is this ok?

    In general, x^x^x^... = Z would imply x = Z^(1/Z)
    But, if x>1, then x^x^x^... is crescent.
    In other words, when "Z" increases, then "x" increases.

    However, lim oo Z^(1/Z) = 1
    So, if x^x^x^... increases, it means that "x" goes to 1 ?!

    Where is the mistake?
     
  2. jcsd
  3. Oct 11, 2007 #2
    x must be between [tex]e^{-e}[/tex] and [tex]e^{\frac{1}{e}}[/tex]. Euler gave a proof showing that it only has a limit in this interval.

    I don't know if the answer is right, but it's in the interval... so it's possible.
     
    Last edited: Oct 11, 2007
  4. Oct 13, 2007 #3
    I reduced it to
    x^k * ln(x) = ln(2) for some k
    so i guess find x and k? like x= 2 and k = 0....
    dunno if this right just wondering.....
     
  5. Oct 13, 2007 #4

    morphism

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    It can be reduced to x^2 = 2 immediately.
     
  6. Oct 13, 2007 #5
    could u explain for some reason I am not able to see it?
    Thanks!
     
  7. Oct 13, 2007 #6
    [tex]x^{x^{x...}}}=x^2=2[/tex]
     
  8. Oct 13, 2007 #7

    HallsofIvy

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    In case you are wondering, Dragonfall is taking x to the power of each side.
    x to the [itex]x^{x^{x...}}}[/itex] is equal to [itex]x^{x^{x...}}}[/itex] since there were an infinite number of "x"s to begin with and according to the equation, that is equal to 2. On the right side, of course, x to the 2 power is x2: 2= x2.
     
  9. Oct 13, 2007 #8
    makes sense overlooked the fact that there are infinite number of them..
    thanks.
     
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