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

1. ## [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

2. ## 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..

3. 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
$k^2 = 2^n$

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.

4. 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