I often use the series of the logarithm to reduce computational effort when computing logarithms. It is usually easier to compute many small logarithms then one large one, so I often exploit this by cutting up the number I want to compute.
This is a simple way to compute $\log x$. We have
$\log x=\log \left(\frac {x}{2}\right)+\log 2$
repeating this process $n$ times with a pre-computed $\log 2$, we have
$\log x=\log \left(\frac{x}{2^{n}}\right)+n\log 2$
We can then quickly compute $\log \left(\frac{x}{2^{n}}\right)$ with a Taylor's Series, its Chebyshev economized counterpart, or a different, more efficient method. I usually choose $n$ as a number that makes $\left(\frac{x}{2^{n}}\right) \approx 1$ to make it easier to estimate in a series.
For example, I may compute $\log 10$ efficiently using the above basic method. Say we know $\log 2$ to $10$ places, i.e. $0.6931471805$. Letting $n=3$ (making $\left(\frac{10}{2^{n}}\right) = 1.25$),
$\log 10=\log \left( \frac{10}{8} \right)+3 * 0.6931471805=\log \left( \frac{5}{4} \right)+2.0794415415$
Already, note that $2.0794415415$ is a decent estimate of $\log 10$. Using 5 terms of the basic Taylor's series expnsion of $\log$, we have
$\log \left( \frac{5}{4} \right)\approx \left( \frac{5}{4} -1\right)-\frac{1}{2}\left( \frac{5}{4} -1\right)^2+\frac{1}{3}\left( \frac{5}{4} -1\right)^3-\frac{1}{4}\left( \frac{5}{4} -1\right)^4+\frac{1}{5}\left( \frac{5}{4} -1\right)^5 \approx 0.22317708\bar{3}$
accurate to 4 decimal places. Together, we have
$\log 10\approx 0.223177083+2.0794415415=2.3026186245$
accurate again to 4 decimal places.