Solving an inequality with terms both within LambertW and outside of it. $\newcommand{\LambertW}{\operatorname{LambertW}}$I am  trying to solve the following inequality:
$$100n^2<2^n, n\in\mathbb{R}$$
I have applied the following steps:
\begin{align}
& \frac{n^2}{2^n} < \frac{1}{100} \\
\Rightarrow {} & n^2 \cdot e^{-\ln(2^n)} <\frac{1}{100} \\
\Rightarrow {} & -n\cdot \ln2 \cdot e^{-n\cdot \ln2} > \frac{-\ln 2}{100n} \\
\Rightarrow {} & -n\cdot \ln2 > W\left(\frac{-\ln2}{100n}\right) \\
\Rightarrow {} & n < \frac{W\left(\frac{-\ln2}{100n}\right)}{-\ln 2}
\end{align}
I do not know where to go from there.
Maple provides me with 3 answers to this solution:
\begin{align}
n_1 & = -2\cdot \LambertW((1/20)\cdot \ln2)/\ln2 \\[8pt]
n_2 & = -2\cdot \LambertW(-(1/20)\cdot \ln2)/\ln2 \\[8pt]
n_3 & = -2\cdot \LambertW(-1, -(1/20)\cdot \ln2)/\ln2
\end{align}
Note that the equality signs mean either $<$ or $>$, because Maple is unable to solve it symbolically using inequalities.
The very last one (being positive); being the correct answer for this specific question.
So how do arrive at these 3 solutions by hand?
 A: So how do arrive at these 3 solutions by hand?

$$n^2\cdot 2^{-n}=\frac{1}{100}\implies$$
$$\left(n^2\cdot 2^{-n}\right)^{1/2}=\left(\frac{1}{100}\right)^{1/2}\implies$$
$$|n|\cdot\exp\left(-\ln2\cdot\frac{n}{2}\right)=\frac{1}{10}\implies$$
$$-\ln(2)\cdot\frac{|n|}{2}\cdot\exp\left(-\ln(2)\cdot\frac{n}{2}\right)=\frac{-\ln(2)}{20}$$
Now, for $|n|=n$, you get:
$$-\ln(2)\cdot \frac{n}{2}=W\left(k,\frac{-\ln(2)}{20}\right),\,\,k\in\mathbb{Z}\implies$$
$$n=-\frac{2W\left(k,\frac{-\ln(2)}{20}\right)}{\ln(2)},\,\,k\in\mathbb{Z}$$
For $|n|=-n$, you get the one with the plus sign as argument:
$$n=-\frac{2W\left(k,\frac{\ln(2)}{20}\right)}{\ln(2)},\,\,k\in\mathbb{Z}$$
It's up to you now to sort the solutions out, depending on your inequality range and the $W$ branch chosen.
A: The function $f:x\mapsto100x^2/2^x$ is decreasing from $+\infty$ to $0$ on $(-\infty,0]$, increasing from $0$ to some value $A$ well above $1$ on $[0,2/\log2]$, and decreasing from $A$ to $0$ on $[2/\log2,+\infty)$. Hence the equation $f(x)=1$ has three roots $x_1\lt0\lt x_2\lt A\lt x_3$ and $f(x)\lt1$ on $(x_1,x_2)\cup(x_3,+\infty)$.
The values $n_i$ in the question are probably the three roots $x_k$. Numerically, $A\approx112.7$, $x_1\approx-.1$, $x_2\approx.1$ and $x_3\approx14.3$.
That $-1\lt x_1\lt0\lt x_2\lt1$ is easy to show by hand, computing $f(-1)$ and $f(1)$. To show that $14\lt x_3\lt15$ by hand, compute some values $f(n)$ with $n$ positive integer. For example, $2^{10}\approx10^3$ hence $f(10)\approx10$ shows that $x_3\gt10$ and $2^{20}\approx10^6\gt1$ shows that $f(20)\approx4\cdot10^{-2}\lt1$ hence $x_3\lt20$. And so on. 
Likewise, pretending that $14^2\approx200$ (since $\sqrt2\approx1.4$) and once again that $2^{10}\approx1000$, one sees that $f(14)\approx100\cdot200/(16\cdot1000)\approx5/4$ hence $x_3$ is definitely near $14$...
Here is an opinion: The usefulness of LambertW in this context is zero, as far as concrete purposes are concerned.
