I noticed that while solving this inequality, I faced a rather annoying fact: the lack of mathematical rigor in the formalism of writing things down.
$$\left| \frac{3}{\ln(\alpha)} \right| < 1$$
Now using the absolute value I get
$$\begin{cases} \phantom{-}\frac{3}{\ln(\alpha)} < 1 & \alpha > 1 \\\\ -\frac{3}{\ln(\alpha)} < 1 & \alpha \in (0, 1) \end{cases}$$
Now the second case is the bitter one, for if one does not pay attention he solves it wrong, indeed one might think
$$-3 < \ln(\alpha) \rightarrow \alpha > e^{-3}$$
Concluding the solution here is $\alpha \in (e^{-3}, 1)$ which is wrong because the correct one is $(0, e^{-3})$.
The fact is that when $\alpha \in (0, 1)$, we have $\ln(\alpha) < 0$. So the "real" inequality reads
$$\frac{-3}{-|{\ln(\alpha)}|} < 1$$
But this creates a redundant absolute value, which again could confuse and most of all a double minus sign, which makes to think it's exactly like the previous case.
How would one take into account the other negative sign, in order to write things properly?
You know $\ln \alpha < 0$ and that multiplying by a negative reverses inequalities, so $-\frac{3}{\ln(\alpha)} < 1$ implies $-3 > \ln(\alpha)$ and so $e^{-3} > \alpha$. I think that's pretty clear.