The Wikipedia page on $\log 2$ mentions a trick to compute $\log 2$ based on the identity
$$2 = \left(\frac{16}{15}\right)^7\left(\frac{81}{80}\right)^3\left(\frac{25}{24}\right)^5$$
It's not too hard to come up with similar-looking formula, for instance:
$$2 = \left(\frac{9}{8}\right)\left(\frac{8}{7}\right)^2\left(\frac{7}{6}\right)^2$$
Consider a formula to have the form
$$2 = \prod_{i=1}^m \left(1+\frac{1}{n_i}\right)^{k_i}$$ where $n_i$ are positive integers and $k_i$ are (possibly negative) integers.
For a given $m$, the best formula maximizes the smallest $n_i$. How would one go about finding such formulas? Is there a trick besides brute force?
Edit: neat, finding consecutive pairs of p-smooth numbers and solving a lattice problem gives us things like
$$2 = \left(\frac{126}{125}\right)^{72}\left(\frac{225}{224}\right)^{27}\left(\frac{2401}{2400}\right)^{-19}\left(\frac{4375}{4374}\right)^{31}$$
and
$$\frac{144}{251} \sum_{k=0}^{\infty} \frac{1}{(2k+1)63001^k} + \frac{54}{449}\sum_{k=0}^{\infty} \frac{1}{(2k+1)201601^k}-\frac{38}{4801}\sum_{k=0}^{\infty} \frac{1}{(2k+1)23049601^k}+\frac{62}{8749}\sum_{k=0}^{\infty} \frac{1}{(2k+1)76545001^k} = \log 2$$
I think I figured it out, thanks to the comment. The key is to pay attention to the prime decomposition of the $n_i$'s and $n_i+1$'s. Let $p$ be the largest prime factor of any integer in $\{n_i\}_i \cup \{n_i+1\}_i$, so that for all i, $n_i$ and $n_{i+1}$ are p-smooth.
Commenter Gerry Myerson usefully points to Stormer's problem and the theorem that, for all $p$ there is a finite set of consecutive integers which are both $p$-smooth.
This yields the following algorithm: fix the value of $p$, find the $p$-smooth pairs of consecutive integers, pick a subset (typically one might need \Pi(p) pairs) and solve the linear system for the values of $k_i$.
This yield the identity mentioned in the edited question.