# [SOLVED] Solving a variable raised to itself

• Apr 5th 2008, 07:19 PM
JesseHackett
[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
• Apr 5th 2008, 07:38 PM
Mathstud28
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..
• Apr 5th 2008, 08:08 PM
mathceleb
Quote:

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.
• Apr 5th 2008, 08:30 PM
mr fantastic
Quote:

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