# Thread: Dealing with a variable raised to itself

1. ## Dealing with 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. ## You cant

that I know of unless you are using the concept of incredibely large numbers....if you are you can say that $\displaystyle x^{x}=2^{n}\approx{x=n\ln(2)}$ otherwise I have no idea.

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
The exact solution is $\displaystyle k = e^{W(\ln 2^n)} = e^{W((\ln 2) n)} \,$, where W is the Lambert W-function.

A search of the MHF will uncover several threads where the Lambert W-function is discussed.

$\displaystyle x^x = a \Rightarrow \ln x^x = \ln a \Rightarrow x \ln x = \ln a \Rightarrow e^{\ln x} \ln x = \ln a \Rightarrow t e^t = \ln a$

where $\displaystyle t = \ln x$.

$\displaystyle t e^t = \ln a \Rightarrow t = W(\ln a)$

by definition of the Lambert W-function

$\displaystyle \Rightarrow \ln x = W(\ln a) \Rightarrow x = e^{W(\ln a)}$.

4. Originally Posted by mr fantastic
The exact solution is $\displaystyle k = e^{W(\ln 2^n)} = e^{W((\ln 2) n)} \,$, where W is the Lambert W-function.

A search of the MHF will uncover several threads where the Lambert W-function is discussed.

$\displaystyle x^x = a \Rightarrow \ln x^x = \ln a \Rightarrow x \ln x = \ln a \Rightarrow e^{\ln x} \ln x = \ln a \Rightarrow t e^t = \ln a$

where $\displaystyle t = \ln x$.

$\displaystyle t e^t = \ln a \Rightarrow t = W(\ln a)$

by definition of the Lambert W-function

$\displaystyle \Rightarrow \ln x = W(\ln a) \Rightarrow x = e^{W(\ln a)}$.
Postscript: An alternative form of this solution is $\displaystyle \frac{\ln a}{W(\ln a)}$.