**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](https://mathoverflow.net/a/116844/163727), 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}



Simple estimation of difference of powers of 2 and powers of 3?
This is about bounding $e^x$ such as $e^x \geq 1 + x$.


For example, when $2^q>3^p$ (ie. $\log\_3 2>\frac{p}q$), from (1.2) we have
$$q\log 2 \geq c\_b \frac{\log 3}{q^{B\_b-1}} + p\log 3,$$
which under exponentiation translates to
$$2^q \geq e^{c\_b \frac{\log 3}{q^{B\_b-1}}}3^p.$$
Then
$$2^q - 3^p \geq \big(e^{c\_b \frac{\log 3}{q^{B\_b-1}}}-1\big)3^p \geq c\_b \frac{\log 3}{q^{B\_b-1}}3^p.$$


The other cases are considered similarly.


