Solving 2^k = n/k for k in Terms of n: Tips and Tricks

[flash]

I would like to solve $$2^k = \frac{n}{k}$$ for k in terms of n, but can't seem to do it. Any help greatly appreciated!

 

[Defennder]

Hm, there may not be a closed form solution. Does anyone else have better luck?

 

[Dick]

No, you can't solve it algebraically. You can only approximate it numerically, unless n is very special.

 

[flash]

Thanks, that's what I wanted to know. Cheers

 

[HallsofIvy]

That depends upon what you mean by "closed form" or "algebraic" solution.

This is obviously equivalent to k2^k= n and, since 2^k= e^kln(2), k e^{k ln(2)}= n. Multiplying on both sides by ln 2, (k ln(2)) e^{k ln(2)}= n ln(2). If we let y= k ln(2), that equation is ye^y= n ln(2).

That equation is directly solvable by the Lambert W function (which is simply defined as the inverse function to f(x)= xe^x): k ln(2)= y= W(n ln(2)) so
k= W(n ln(2))/ln(2).