An accepted answer is in the cross post at https://mathoverflow.net/questions/428396/simple-estimation-of-difference-of-powers-of-2-and-powers-of-3 .
1. Question
How to get from the formulas
$$ \left| \frac{\log 2}{\log 3} - \frac{p}{q} \right| < c_a\frac{1}{q^{B_a}} \ \ \ \ \ \ \ \ \ \ \ (1.1)$$
and / or
$$ \left| \frac{\log 2}{\log 3} - \frac{p}{q} \right| \geq c_b\frac{1}{q^{B_b}} \ \ \ \ \ \ \ \ \ \ \ (1.2)$$
to the formula
$$ \left| 3^p - 2^q \right| \geq c_r\frac{1}{q^{B_r}}3^p\ \ \ \ \ \ \ \ \ \ \ \ (1.3)$$
with constants $c_a, c_b, B_a, B_b \in \mathbb{R}$ given and $\frac{p}{q} \in \mathbb{Q}$?
It is also important for me how exactly $c_r, B_r \in \mathbb{R}$ depend on the other constants $c_a, c_b, B_a$, and $B_b$. All 6 constants are > 0 and $B_a, B_b \in [2, 8)$.
The lower bound in (1.3) is claimed by https://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/ (proposition 3 and corollary 4) . Terry Tao uses (1.2) with $\geq$ instead of (1.1) with $<$. Maybe both inequalities are required to prove (1.3) for the two cases $3^p > 2^q$ and $3^p < 2^q$.
Something very similar is the answer by user Lierre to the MO question distance between powers of 2 and powers of 3, but I would like to see more detailed steps and how to calculate the constants in (1.3).
2. Ansatz A
\begin{align} 3^p - 2^q &= 3^p - (3^{\log_3(2)})^q\\ &= 3^p - 3^{q\cdot \log_3(2)}\\ &= 3^p - 3^{q\cdot \log_3(2) - p + p}\\ &= 3^p - 3^p\cdot3^{q\cdot \log_3(2) - p}\\ &= 3^p - 3^p\cdot3^{q\cdot(\log_3(2) - \frac{p}{q})}\\ &= 3^p(1 - 3^{q\cdot(\log_3(2) - \frac{p}{q})}) \ \ \ \ \ \ \ \ \ \ (2.1)\\ \end{align}
Note that
\begin{align} 3^p > 2^q &\iff \log_3(3^p) > \log_3(2^q)\\ &\iff p > q\log_3(2)\\ &\iff \frac{p}{q} > \log_3(2)\\ &\iff 0 > \log_3(2) - \frac{p}{q}\\ &\iff \log_3(2) - \frac{p}{q} < 0 \ \ \ \ \ \ \ \ \ \ (2.2)\\ \end{align}
and analogously
$$2^q > 3^p \iff \log_3(2) - \frac{p}{q} > 0 \ \ \ \ \ \ \ \ \ \ (2.3)$$.
3. Ansatz B
\begin{align} 3^p - 2^q &= 3^p(1 - \frac{2^q}{3^p})\\ &= 3^p(1 - \frac{(e^{\log2})^q}{(e^{\log3})^p})\\ &= 3^p(1 - \frac{e^{q\log2}}{e^{p\log3}})\\ &= 3^p(1 - e^{q\log2 - p\log3}) \ \ \ \ \ \ \ \ \ \ (3.1)\\ \end{align}
The logarithms of $2^n$ and $3^m$ can be arbitrarily close together for n, m > 0. That doesn’t mean the numbers themselves get close.