I'm playing around with some math and I arrived at the equation:
$$\left ( \frac{x}{3}\right )^{n} e^{- \left ( x/3\right )^{n} } = \text{const.}$$
where I'm trying to solve for $x$. I think I can solve this using the Lambert W function, but I'm not sure how to apply it when the exponential term contains $x$ raised to the power $n$. Can I just apply the approach used in this example? Or is there something I'm missing.
Also, $const.$ indicates an odd-looking constant I chose to abbreviate.
Starting with
$$\left(\frac{x}{3}\right)^n e^{-(x/3)^n} = [\mathrm{const.}]$$
first note that the exponent of $e$ contains a negative sign. We can produce that on the outside by negating:
$$-\left(\frac{x}{3}\right)^n e^{-(x/3)^n} = -[\mathrm{const.}]$$
and then we see by grouping, that
$$\left[-\left(\frac{x}{3}\right)^n\right] e^{\left[-\left(\frac{x}{3}\right)^n\right]} = -[\mathrm{const.}]$$
Now that we have the same quantity both outside and inside the exponential, we can take the W:
$$-\left(\frac{x}{3}\right)^n = W\left(-[\mathrm{const.}]\right)$$
hence with another negation,
$$\left(\frac{x}{3}\right)^n = -W\left(-[\mathrm{const.}]\right)$$
and now the $n$th root
$$\frac{x}{3} = \sqrt[n]{-W\left(-[\mathrm{const.}]\right)}$$
hence, finally
$$x = 3 \sqrt[n]{-W\left(-[\mathrm{const.}]\right)}$$
is the solution.