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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Originally Posted by JesseHackett 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 Solving for k, we take the square root of both sides: 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.
Originally Posted by JesseHackett 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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