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