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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- April 5th 2008, 08:19 PMJesseHackett[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 - April 5th 2008, 08:38 PMMathstud28Just 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..

- April 5th 2008, 09:08 PMmathceleb
- April 5th 2008, 09:30 PMmr fantastic
Answered in this thrread: http://www.mathhelpforum.com/math-he...ed-itself.html