Results 1 to 4 of 4

Math Help - Dealing with a variable raised to itself

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    3

    Dealing with a variable raised to itself

    I am trying to solve for k and i get to the step:

    2^n=k^k

    How do i deal with k raised to itself....its driving me nuts
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    You cant

    that I know of unless you are using the concept of incredibely large numbers....if you are you can say that x^{x}=2^{n}\approx{x=n\ln(2)} otherwise I have no idea.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by JesseHackett View Post
    I am trying to solve for k and i get to the step:

    2^n=k^k

    How do i deal with k raised to itself....its driving me nuts
    The exact solution is k = e^{W(\ln 2^n)} = e^{W((\ln 2) n)} \, , where W is the Lambert W-function.

    A search of the MHF will uncover several threads where the Lambert W-function is discussed.


    x^x = a \Rightarrow \ln x^x = \ln a \Rightarrow x \ln x = \ln a \Rightarrow e^{\ln x} \ln x = \ln a \Rightarrow t e^t = \ln a

    where t = \ln x.

    t e^t = \ln a \Rightarrow t = W(\ln a)

    by definition of the Lambert W-function

     \Rightarrow \ln x = W(\ln a) \Rightarrow x = e^{W(\ln a)}.
    Last edited by mr fantastic; April 5th 2008 at 07:53 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    The exact solution is k = e^{W(\ln 2^n)} = e^{W((\ln 2) n)} \, , where W is the Lambert W-function.

    A search of the MHF will uncover several threads where the Lambert W-function is discussed.


    x^x = a \Rightarrow \ln x^x = \ln a \Rightarrow x \ln x = \ln a \Rightarrow e^{\ln x} \ln x = \ln a \Rightarrow t e^t = \ln a

    where t = \ln x.

    t e^t = \ln a \Rightarrow t = W(\ln a)

    by definition of the Lambert W-function

     \Rightarrow \ln x = W(\ln a) \Rightarrow x = e^{W(\ln a)}.
    Postscript: An alternative form of this solution is \frac{\ln a}{W(\ln a)}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 28th 2011, 01:35 PM
  2. Replies: 2
    Last Post: May 1st 2011, 08:03 AM
  3. Replies: 1
    Last Post: February 24th 2009, 08:57 PM
  4. any no raised to zero
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: April 19th 2008, 01:16 PM
  5. [SOLVED] Solving a variable raised to itself
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 5th 2008, 08:30 PM

Search Tags


/mathhelpforum @mathhelpforum