How to compute an upper bound for k in $n(\frac{en}{km})^k = \frac{1}{10}$

Question

I'm wondering how to compute $k$ in $n(\frac{en}{km})^k = \frac{1}{10}$ or at least give an upper bound. $e$ is Euler's number. $n$, $m$ and $k$ are positive integers.

Answer 1

This can be done using the Lambert W function, the unique number satisfying

$$x = W(x)e^{W(x)}$$

We can simplify the equation into

$$10n\left(\frac{en}{m}\right)^k = k^k$$

Let $10n = a, \frac{en}{m}=b$:

$$a(b)^k = k^k$$

$$b(a^{\frac{1}{k}}) = k$$

Taking the logarithm of both sides:

$$\log b + \frac{1}{k} \log a = \log k$$

Letting $k=e^x$:

$$\log b +(\log a)e^{-x} = x$$

$$(\log a)e^{-x} = x-\log b$$

Multiplying by $e^x$:

$$\log a = (x-\log b)e^x$$

Dividing by $b$:

$$\frac{\log a}{b} = (x-\log b)e^{(x-\log b)}$$

$$W\left(\frac{\log a}{b}\right) = x-\log b$$

$$\log b + W\left(\frac{\log a}{b}\right) = x$$

$$\log\left(\frac{en}{m}\right) + W\left(\frac{\log(10n)}{\frac{en}{m}}\right) = \log(k)$$

$$k = \left(\frac{en}{m}\right)\exp(\left(W\left(\frac{m\log(10n)}{en}\right)\right)$$