Consider the equation $9x^2e^{-x}=1$, Let c be the greatest positive root. Find the greatest positive integer $n$ such that $n<c$.
I was able to do this question by converting it into $9x^2=e^x$, then into $2\ln(3)+2\ln(x)-x=0$, then I defined $f(x)=2\ln(3)+2\ln(x)-x$ and showed that there is a sign change between $f(5)$ and $f(6)$ by using rough decimal approximations that I happened to know. This leads to the answer of $n=5$. I should note that a previous part of the question had me sketch the graph $y=9x^2e^{-x}$ for some intuition.
My problem is the fact that I had to use knowledge of decimal expansions which is not a favourable method for me. I'm curious to see if anyone knows of more efficient or elegant methods to answer this problem.
Thanks in advance.
Sooner or later, you will learn that the only explicit solutions of the equation are given in terms of Lambert function which is very useful and extensively used in many areas. Just by curiosity, type Lambert on the eseach bar of the site and you will notice $4047$ entries.
The root you are looking for is given by $$x=-2 W_{-1}\left(-\frac{1}{6}\right)$$ which you can approximate using the expansion given in the linked page
$$W_{-1}(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where where $L_1=\log(-t)$ and $L_2=\log(-L_1)$.
Using the truncated expansion, this would give $x \sim 5.70664$ while the exact solution is $x=5.66630$.