Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
The following general inequality relating $k$ and $q$ is proved in this question: $$k \geq \log_{5}{\bigg(\frac{q}{q-4}\bigg)} \tag{1}.$$
On the other hand, Dris proved in 2012 (see Journal of Integer Sequences) the inequality $$q^k < \frac{2n^2}{3}$$ which implies that $$k < \log_{q} 2 - \log_{q} 3 + 2\log_{q} n. \tag{2}$$
Together, Inequalities (1) and (2) imply that $$\log_{5} q - \log_{5} (q - 4) \leq k < \log_{q} 2 - \log_{q} 3 + 2\log_{q} n,$$ so that we have $$\log_{5} q - \log_{5} (q - 4) < \log_{q} 2 - \log_{q} 3 + 2\log_{q} n. \tag{3}$$
Here is my:
QUESTION: Can Inequality (3) be solved for $q$ in terms of $n$ (or the other way around), if $q^k n^2$ is an odd perfect number with special prime $q$? If YES, can you show how? If NO, can you explain why?
MY ATTEMPT
I tried to ask WolframAlpha for the solutions of Inequality (3) to $\log q$ and did also try to ask WolframAlpha for the solutions of Inequality (3) to $\log n$, but I do not know how to interpret the results.
In particular, solving for $\log n$ from Inequality (3), WolframAlpha is giving me $$\Bigg(\log n > \frac{\log(q)\log\bigg(\frac{q}{q-4}\bigg) - (\log(2) - \log(3))\log(5)}{2\log(5)}\Bigg) \land \Bigg(\log\bigg(\frac{q}{q-4}\bigg) \in \mathbb{R}\Bigg) \land \Bigg(\log(q) > 0\Bigg),$$ where we know that the second and third conditions are always satisfied since $q$ being the special prime implies that $q \geq 5$.
Alas, this is where I get stuck!