I did this with my students a while ago. First I got them to construct their own slide rules using Briggs estimation technique. Then a slightly more accurate table translating between base 10 logs, decibels, base 2 logs, and musical notes ($semitone^{12}=2$).
After that we studied Briggs' methods to improve the estimates and create accurate logs.
My source material for Napier is http://profmarino.it/Nepero/napier1619construction.pdf
My source material for Briggs is a translation of Chapter 7 of his Arithmetica Logrithmica http://www.17centurymaths.com/contents/albriggs.html I have since found a more easily understood document explaining his techniques https://hal.inria.fr/inria-00543939/document
Napier
If I remember correctly, Napier's method was really weird. He started with a big number and basically multiplied by something like 0.999999 to get lower numbers. He did an insane number of calculations at up to 32 bit precision, so he only got parts of his table finished.
Something to note, logs are about ratios, not numbers, so the base number of a log table is arbitrary. We set the base of our ratio to be 1 to make them trivial to calculate, so $\log 1 =0$ ... Napier did not :-(
Briggs method for initial estimation
Henry Briggs decided that there had to be a better way and used a number of techniques around base 10, with the base ratio set to 1. In Napier's system it was annoying to multiply or divide numbers by 10 so you could use the parts of the log tables Napier had completed, but with $\log 10=1$ and $\log 1=0$ it is trivial. This not only cuts down on an extra calculation but reduces the accumulation of rounding errors.
The following describes how Briggs did this hundreds of years ago by hand, with much greater precision than a calculator.
As a first estimate, observe that
$$2^{10} \approx 10^3$$
Therefore we can calculate
$$10\log 2 \approx 3\log 10$$
$$\log 2 \approx 0.3$$
Briggs method for calculating correction factor
But Briggs needed to improve on that. The correction factor is $1.024$. If we find out the log of that, we can add it into the above equation. We will use 3 facts to estimate this number. First, since ancient times there has been an easy algorithm for finding square roots (for my students, it is called "the square root button"), and repeated application takes numbers close to $1$. Second, the log of a square root is exactly half the log of original number. Third, we have set the log of 1 to be 0. Therefore all we need to do is find a root of 10 close to 1, and then we can easily interpolate to accurately find more logs.
To find a root of 10 close to 0, take repeated square roots until you get a small number, say
$$ \sqrt{\sqrt{\sqrt{\sqrt{{\sqrt{\sqrt{\sqrt{\sqrt{10}}}}}}}}} = 10^{(\frac 1 2 ^8)} = 1.0090$$ $$\log 1.0090 = (\frac 1 2 )^8$$
The interval between $\log 1$ and $\log 1.0090$ is linear up to 4dp. Using smaller numbers rapidly improves accuracy, but this is sufficient for a class project. Briggs used 54 square roots, accurate to 32 dp.!
Next we do the same to 1.024 to get a fraction close to $1.0090$, we'll use $\sqrt {1.024} = 1.0119$. We use extrapolation between
$\log 1 = 0 $ and
$\log 1.0090 = \frac {1}{256}$
The interpolation formula is simplified as $y_0=0$
$$ y = y_1 \times \frac {x -x_0}{x_1-x_0}$$
$$\log 1.0119= \log 1.0090 \times \frac {0.0119}{0.0090} = 0.00516$$ $$\log 1.024 = 2 \log 1.0119 = 0.0103$$
Putting it all together
We then use this correction factor to get good precision based on our initial estimate
$$ 2^{10} = 10^3 * 1.024$$
$$ 10 \log 2 = 3 + 0.0103$$
$$ \log 2 = 0.30103$$
$$ 10^{0.30103}=2.00000...$$
The technique itself magnifies the accuracy. The next to use is $6^9 = 10^7 * 1.0077696$
