The "efficiency" of the set of denominators $\lbrace 31, 49, 161 \rbrace$ is $0.744$. It is better to use $\lbrace 251, 449, 4801, 8749\rbrace$, which lets you compute the logs of the first $4$ primes for an efficiency of $0.573$. Those are the largest numbers I could find which are sandwiched between $7$-smooth numbers. Using the first $5$, $6$, or $7$ primes didn't improve this, as they resulted in slightly larger efficiencies, e.g. $0.601$ for $\lbrace 28799,57121,62425,74359,246401,388961,672281 \rbrace$, the $7$ largest numbers I could find so that adding or subtracting $1$ produces a $17$-smooth number. 

$$\begin{eqnarray} \operatorname{arctanh}\frac{1}{251} &=& \frac{1}{2}(\log 2 + 2 \log 3 - 3 \log 5 + \log 7) \newline
\operatorname{arctanh}\frac{1}{449} &=& \frac{1}{2}(-5\log 2 + 2 \log 3 + 2 \log 5 - \log 7) \newline
\operatorname{arctanh}\frac{1}{4801} &=& \frac{1}{2}(-5\log 2 -\log 3 - 2\log 5 + 4 \log 7) \newline
\operatorname{arctanh}\frac{1}{8749} &=& \frac{1}{2}(-\log 2 -7\log 3+4\log 5+\log 7). \end{eqnarray}$$

By inverting this system we get

$$\begin{eqnarray}\log 2 &=& 144~a(251) + 54~a(449)-38~a(4801)+62~a(8749)  \newline \log3&=& 228~a(251)+86~a(449)-60~a(4801)+98~a(8749) \newline\log5 &=& 334~a(251) + 126~a(449)-88~a(4801)+144~a(8749)  \newline \log 7 &=& 404~a(251)+152~a(449)-106~a(4801)+174~a(8749) \end{eqnarray}$$


where $a(n) = \operatorname{arctanh}\frac{1}{n}$.


It is plausible that it would be better to use a small set of primes, but not the smallest ones.