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Henry Briggs compiled the first table of base-$10$ logarithms in 1617, with the help of John Napier. My question is: how did he calculate these logarithms? How were logarithms calculated back then?

I've found these pages to be fairly useful, but they don't seem to say much in the way of what I'm looking for. Any answers or useful references would be appreciated.

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    $\begingroup$ Well, you just hammer down a few logs and a plate, and you get a table. :-) $\endgroup$ – Asaf Karagila Sep 5 '13 at 17:14
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    $\begingroup$ @AsafKaragila Naturally ... $\endgroup$ – Hagen von Eitzen Sep 5 '13 at 17:20
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    $\begingroup$ @HagenvonEitzen if he did it naturally, it would be a natural-log table :P $\endgroup$ – Omnomnomnom Sep 5 '13 at 17:29
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    $\begingroup$ @exk an interesting proposition, given that Brook Taylor introduced Taylor series 100 years later. To be sure though, the Mercator series was first published in 1668, and isn't a terribly efficient means of calculation except for values sufficiently near $1$. $\endgroup$ – Omnomnomnom Sep 5 '13 at 18:22
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    $\begingroup$ Richard Feynman gives an account of Briggs's work in Chapter 22, "Algebra" of "The Feynman Lectures on Physics, Volume 1". $\endgroup$ – awkward Sep 5 '13 at 23:14
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At MAA you may find 'Logarithms: The early history of a familiar function' while Napier's logarithms are described with care in Roegel's article 'Napier's ideal construction of the logarithms' (rather less nice than the usual ones since using $10^7$ as reference!). A shorter description was given by Lexa in 'Remembering John Napier and His Logarithms' and should provide a good starting point.

Napier's work itself appears in 'A Description of the Admirable Table of Logarithms' : Edward Wright's $1616$ translation of Napier's Latin book.
A book from $1915$ named 'Napier tercentenary memorial volume' is proposed by archive.org.

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You can find the details in e: The Story of a Number. The basic idea is that square roots are easy to calculate. If you want for example $\log_{10}2$ (the number such $2=10^{\log_{10}2}$): $$10^{0.25} = 10^{1/4} = 1.778...< 2 < 3.162... = 10^{1/2} = 10^{0.5},$$ i.e., $$0.25 < \log_{10}2 < 0.5$$ and multiplying/dividing for $10^{1/2^k}=\sqrt{\sqrt{\cdots\sqrt{10}}}$ you can get better approximations.

Also important: the successive square roots of 10 are calculated once and can be used many times.

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  • $\begingroup$ multiplying and dividing by the square and higher roots of 10 can get very ugly very fast when working to 40 digits. (Euler was a wiz at square roots.) It would be much easier if you could divide by 1.1111111, 1.01010101, 1.001001001, etc. Now the division is simple! Shift and subtract. You can do it on an abacus, though two might be handier. The table of logarithms is greatly reduced. I think that Napier's original log table was not a log table at all, but was a table of powers of 1.0000001... created by shift and add. $\endgroup$ – richard1941 Aug 26 '16 at 5:37
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This video from numberphile could be worth a watch

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  • $\begingroup$ That's precisely what prompted the question :) $\endgroup$ – Omnomnomnom Sep 6 '13 at 23:05
  • $\begingroup$ What a bad idea to suggest this... $\endgroup$ – Did Nov 1 '16 at 15:29

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