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Solve (x)^x^3=3 find X

  1. Mar 9, 2008 #1
    Hello there

    I've seen this question and tried to solve it but until now for no avail., I couldn't solve it but I know that the value of X is cubic root of 3

    [​IMG]

    Thanks.
     
  2. jcsd
  3. Mar 9, 2008 #2
    i dont think you can solve that analytically, instad you need to use numerical method?
     
  4. Mar 9, 2008 #3
    Someone solved it by Calculus.
     
  5. Mar 9, 2008 #4

    HallsofIvy

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    Do you have a reference for that? It certainly can be solved by "inspection"- noting that
    [tex]\sqrt{3}^{\sqrt{3}^3}= \sqrt{3}^3= 3[/itex].

    But I have no idea what you mean by "solved it by Calculus"!
     
  6. Mar 9, 2008 #5
    I didn't understand how he solved it but I've seen Transcendental Functions, he derived something, I recall that there was Inverses and then he got e^1/3, if you still have no idea, tell me and I'll bring the solution from him.
     
  7. Mar 9, 2008 #6

    D H

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    Surely you mean
    [tex]\sqrt[3] 3^{\sqrt[3] 3^3} = \sqrt[3] 3^3 = 3[/tex]
     
    Last edited: Mar 9, 2008
  8. Mar 9, 2008 #7

    HallsofIvy

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    Yes, sorry. Very careless of me.
     
  9. Mar 9, 2008 #8
    Seems easy to generalize:

    [tex]x^{x^n}=n[/tex]
    [tex]\sqrt[n] n^{\sqrt[n] n \ ^n} = \sqrt[n] n \ ^n = n[/tex]
     
    Last edited: Mar 9, 2008
  10. Mar 9, 2008 #9

    D H

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    An even more interesting generalization is

    [tex]\underbrace{\sqrt[n]^n^{\sqrt[n]^n^{\dotsm^{\sqrt[n]^n}}}}_{n+1} = n[/tex]
     
  11. Mar 9, 2008 #10
    Why is it n+1 levels?
     
  12. Mar 9, 2008 #11
    Not sure if this is interesting but changin the original problem somewhat:

    [tex]x^{x^n}=m[/tex]

    And assuming a solution of the form:

    [tex]x= \sqrt[n]{nq}[/tex]

    We find that q is determined as follows:

    [tex](nq)^q=m[/tex]
     
  13. Mar 9, 2008 #12
    it doesn't seem to matter at all how many levels you take it. for example

    [tex] \sqrt [5]^5^{(\sqrt [5]^5)}^{5}=5[/tex]
     
  14. Mar 10, 2008 #13

    D H

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    That is just a specific instance of John's generalization:
    With [itex]n[/itex] as the final power it does not matter how many levels you take it. Start with the tautology:

    [tex]\sqrt[n] n \ ^n = n[/tex]

    Substituting [itex]\sqrt[n] n ^n[/itex] for any of the ns on the left hand side of the tautology will yield a valid expression. Substituting for the power yields

    [tex]\sqrt[n] n^{\sqrt[n] n \ ^n}= n[/tex]

    This substitution can be carried on indefinitely:

    [tex]\sqrt[n] n ^{\sqrt[n] n ^{\cdots^{\sqrt[n] n \ ^n}}}= n[/tex]

    My "more interesting" generalization (post #9) replaces the final power ([itex]n[/itex]) with with [itex]\sqrt[n] n[/itex]. Changing the final power from [itex]n[/itex] to [itex]\sqrt[n] n[/itex] like I did in post #9 yields expressions like

    [tex]\sqrt 2^{\sqrt 2^{\sqrt 2}} = 2[/tex]

    and

    [tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = 3[/tex]
     
    Last edited: Mar 10, 2008
  15. Mar 10, 2008 #14
    How do you show that?
     
  16. Mar 10, 2008 #15
    Don't you think this should read like this:

    [tex]\sqrt[3] 3^{\sqrt[3] 3^{\sqrt[3]3^{\sqrt[3]3}}} = \sqrt [3]^3[/tex]
     
  17. Mar 10, 2008 #16

    D H

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    Apologies first: I should have used parentheses. Without parentheses, exponentiation evaluates right-to-left. Post #9 is correctly written as

    [tex]\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1} = n[/tex]

    Taking the log (base n) of the left hand side,

    [tex]\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =
    \sqrt[n] n \,
    \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr) =
    {\sqrt[n] n}^{\;2} \,
    \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr) =
    \cdots =
    {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right) = n\cdot \frac 1 n = 1
    [/tex]
     
  18. Mar 10, 2008 #17

    WOW, nice trick!!!!!
     
  19. Mar 10, 2008 #18
    Given where you placed the parenthesis, I still don’t follow.
     
  20. Mar 10, 2008 #19

    D H

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    Step by step, then. The problem is to evaluate

    [tex]\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}[/tex]

    To do so I will take the base-n log of this expression. Using [itex]\log a^b = b\log a[/itex]

    [tex]\log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1}\Biggr) =
    \sqrt[n] n \,
    \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_n\Biggr)
    [/tex]

    We still have something of the form [itex]\log a^v[/itex]. Applying the identity again yields

    [tex]
    {\sqrt[n] n}^{\;2} \,
    \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n-1}\Biggr) [/tex]

    After doing this [itex]m<n[/itex] times, the expression becomes

    [tex]
    {\sqrt[n] n}^{\;m} \,
    \log_n\Biggl(\underbrace{\sqrt[n] n^{\left(\sqrt[n] n^{\left(\dotsm^{\sqrt[n] n}}\right)}\right)}_{n+1-m}\Biggr)
    [/tex]

    After applying this identity n times, the log (base n) of the original expression becomes

    [tex]
    {\sqrt[n] n}^{\;n} \log_n \left(\sqrt[n] n\right)
    [/tex]

    The first term, [itex]{\sqrt[n] n}^{\;n}[/itex] is [itex]n[/itex]. The second term, [itex]\log_n \left(\sqrt[n] n\right)[/itex] is [itex]1/n[/itex]. The product of these two terms is one. If the log (base n) of the expression in question is one, the expression itself must be equal to n.
     
  21. Mar 10, 2008 #20
    I meant that it was solved using Calculus, I got the solution and I uploaded it.

    [​IMG]
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    [​IMG]
     
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