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X^x^x^x^x = 2

  1. Aug 18, 2010 #1
    In a book of math puzzles Peter Winkler discusses the sequence x, x^x, x^x^x, x^x^x^x and writes about conditions for its divergence. Clearly for x=1 the sequence is 1,and for x=2 it diverges to infinity.

    Then he shows that the maximum value of x for the sequence to converge is x=e^(1/e) or
    x= 1.44467... At this value the infinite tower of "x" exponents is equal to e (2.7182818..).
    For any x larger than e^(1/e), the sequence diverges to infinity.

    Mr Winkler later goes on to discuss the equation x^x^x^x^x...=2 (an infinite tower of "x"
    exponents=2)
    By using the trick that the exponent of the bottom"x" is the same as the whole expression,the equation becomes x^2=2, and x=sqrt(2)=1.414... is the solution.
    (This is close to the maximum value for convegence( shown above) 1.44467...

    My question(finally):
    Suppose you have an equation x^x^x^x^x^x^x...=10 (an infinite tower of "x"=10)
    Why can't you use the same trick as we did for x^x^x^x^x^x=2 case.
    In this case we would get x^10=10. And the solution is x= the tenth root of 10(x=1.2589..)
    Now the tenth root of 10 is clearly the wrong answer because:
    1) It is too small . It is smaller than sqrt(2) whose tower conveges to the number 2
    2) The maximum value this tower of "x" converges to is 2.71828... at x=e^(1/e)

    Why doesn't this trick work for x^x^x^x^x^x...=10 ?
     
  2. jcsd
  3. Aug 18, 2010 #2
    Re: x^x^x^x^x...=2

    Because there is no x for which the sequence converges to 10? The trick works for small x because we presume that the solution exists, which is not the case here.
     
  4. Aug 18, 2010 #3
    Re: x^x^x^x^x...=2

    Thanks hamster. Sometimes you can forget, some equations have no solutions.
     
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