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Problem on exponent

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data
    x^X^x^x^x=2. find value of x.


    2. Relevant equations

    taking log both sides, but it makes a equation which i am not able to solve.

    3. The attempt at a solution
    x^(x)^4=2
    x^4logx=log2. what next??????
     
  2. jcsd
  3. Mar 25, 2013 #2

    Mentallic

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    Is that

    [tex]x^{x^{x^{x^x}}}[/tex]

    or

    [tex](((x^x)^x)^x)^x[/tex]

    ?
     
  4. Mar 25, 2013 #3
    the fist one
     
  5. Mar 25, 2013 #4
    the first one
     
  6. Mar 25, 2013 #5

    Mentallic

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    Was this a problem you just made up? In any case, I don't believe you'll be able to solve it using analytic functions.

    If you take the log of both sides, it becomes

    [tex]x^{x^{x^x}}\log x = \log 2[/tex]

    and not [itex]x^4\log x[/itex] as you suggested. Also, the power tower isn't

    [tex]x^{x^4}[/tex] either. If you use Knuth's up arrow notation, then it would be equivalent to x^^5 where each ^ represents an up arrow.
     
  7. Mar 25, 2013 #6
    then what should i do how to solve
     
  8. Mar 25, 2013 #7

    Curious3141

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    Funnily enough, you can solve the infinite power tower quite easily, provided a solution exists, e.g.

    [tex]x^{x^{x^{x^{x}...}}} = 2 \implies x^2 = 2 \implies x = \sqrt{2}[/tex]

    But there's no way of solving a finite power tower equation algebraically. You can probably do it numerically, though, to get an approximate solution.

    EDIT: you can use ##x = \sqrt{2}## as an excellent starting guess for a numerical solver for your finite power tower problem.
     
    Last edited: Mar 26, 2013
  9. Mar 25, 2013 #8

    Mentallic

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    Last edited by a moderator: May 6, 2017
  10. Mar 26, 2013 #9
    Curious3141 , I am curious to know how x^x^x .... = 2 became x^2 = 2 .
     
  11. Mar 26, 2013 #10

    Mentallic

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    If x^x^x... = 2 then if we add another x at the bottom of that tower power, we haven't changed anything since the power tower is infinitely high. Or in other words,
    y^x^x^x... = 2 where y=x is still equivalent to x^x^x..., so we substitute 2 for x^x^x... since that's what we assumed it is equal to, and arrive at

    y^2 = 2
    y=x
    x^2 = 2

    Curious tends to instigate these feelings in all of us :smile:
     
    Last edited: Mar 26, 2013
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