Best approximation of log 3?

One way to use the series you have is to observe that $\log x =-\log \frac 1x$

Given any positive number $x$ we can define

$u=(x-1)/(x+1)$

so that $x=(1+u)/(1-u)$. Then

$\log x = \log (1+u)-\log(1-u)$

where the Taylor series converges for both terms on the right side because $-1<u<+1$.

One advantage of this method is that for a given value of $x$ the series converges faster than using $\log x =-\log(1/x)$, because the series argument is smaller in size.. For example, with $\log3$ we have $u=1/2$, whereas for $\log(1/3)$ we would have to use the series to compute $\log(1-2/3)$.