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Math Help - [SOLVED] Solving a variable raised to itself

  1. #1
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    [SOLVED] Solving 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
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Just to let you know

    You are not supposed to double post...you can get in trouble for it...if you leave your question in one forum it will be answered eventually..
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  3. #3
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    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
    k^2 = 2^n

    Solving for k, we take the square root of both sides:

    \sqrt{k^2} = \sqrt{2^n}

    k = \pm  2^{n/2}

    EDIT: I just saw you had k raised to itself, and not 2. Disregard my post.
    Last edited by mathceleb; April 6th 2008 at 08:14 AM.
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  4. #4
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    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
    Answered in this thrread: http://www.mathhelpforum.com/math-he...ed-itself.html
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