TITLE: Solving $1.1^n = n^{100}$
QUESTION [6 upvotes]: How do I go about solving for $n$ in the following equation:
$$(1.1)^n = n^{100}$$
A hint suffices.

REPLY [2 votes]: First you can rewrite the equation a bit, as
$$\begin{align}1.1^n &= n^{100} \\ 1.1^{n/100} &= n \\ (1.1^{1/100})^n &= n \\ p^n &= n & (\text{where } p = \sqrt[100]{1.1}).\end{align}$$
This is where you normally get stuck, but by introducing the somewhat common notation of the Lambert W function $W$ satisfying $z = W(z) e^{W(z)}$ you can "cheat" your way to a closed-form expression for $n$. Example 1 on the Wikipedia page shows how to get to the following solution:
$$n = \frac{W(-\log a)}{-\log a}$$
If you insist, you can find numerics in Mathematica with the command ProductLog, which is just a different name for the Lambert W function. Feeding it the above, you get the solutions $1.00095$ and $9623.25$.