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+
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..
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 $\displaystyle k^2 = 2^n$ Solving for k, we take the square root of both sides: $\displaystyle \sqrt{k^2} = \sqrt{2^n}$ $\displaystyle k = \pm 2^{n/2}$ EDIT: I just saw you had k raised to itself, and not 2. Disregard my post.
Last edited by mathceleb; Apr 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
View Tag Cloud