Dealing with a variable raised to itself

**JesseHackett** · April 5th 2008, 07:13 PM

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

**Mathstud28** · April 5th 2008, 07:18 PM

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

**mr fantastic** · April 5th 2008, 07:38 PM

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

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

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

where $t = \ln x$ .

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

by definition of the Lambert W-function

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

**mr fantastic** · April 5th 2008, 08:24 PM

Originally Posted by mr fantastic

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

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

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

where $t = \ln x$ .

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

by definition of the Lambert W-function

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

Postscript: An alternative form of this solution is $\frac{\ln a}{W(\ln a)}$ .

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