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I'm teaching an Algebra II/Trigonometry class that is currently working on logarithms as the inverse function of exponential functions. Now, to keep their interest I had them struggle for a few minutes with $2^x = 10$ and trying to solve it. Trying to find the answer to that question is the driving force behind our activities with logarithms (thankfully my students really want an answer to that question, so they're willing to learn whatever I have for them so long as they can get closer to answering the question posed originally.

Some of them have noticed that there is a "log" function on their calculator, which I acknowledged, but I also warned them that right now if they plugged 10 into their calculator they wouldn't get the correct answer.

I want to show them tomorrow as a quick history lesson before some presentations that before we had calculators that could calculate logarithms, there were tables of logarithms that people kept in books.

I know that they're going to ask, and as of right now my searches haven't yielded an answer:

How were the values in the old logarithm tables found? I've found a lot of information on how people used logarithm tables, but very little information on how those values were calculated. Can anyone help me here?

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  • $\begingroup$ As far as $2^x = 10$ goes, It can be solve using my method discussed in the question Method of Solving 5x=326 (Logs not allowed). $\endgroup$ – JTP - Apologise to Monica Mar 12 '15 at 2:22
  • $\begingroup$ Well from that link I've got a tentative answer, but if someone is able to give me more information I'd definitely love to hear it. $\endgroup$ – Jhecht Mar 12 '15 at 2:29
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    $\begingroup$ And the calculus method was using the approximation of roots for equations like $2^x = 100$ being rearranged to $2^x - 100 = 0$ by the Newton Method and linear approximation? $\endgroup$ – Jhecht Mar 12 '15 at 2:55
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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 $6^9 = 10^7 * 1.0077696$

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  • $\begingroup$ I thought Napier used something like powers of 1.000001. In any case, I appreciate the link to Roegel's analysis of the methods Briggs seems to have used. Gerhard "Still Likes The Geometric Construction" Paseman, 2015.03.13 $\endgroup$ – Gerhard Paseman Mar 13 '15 at 17:51
  • $\begingroup$ @GerhardPaseman Michael E2's link in the comments shows one of his tables. He starts at log 10000000 = 0 and effectively multiplied by 0.9999999 each time so log 9999999 = 1. It would still be a simple subtraction to calculate each term. I don't properly understand Napier myself, as arithmetic concepts were so different back then (decimals were a new thing), but I have read several misinterpretations of his work that try to shoehorn what he did into modern models. In the end his understanding was so different I couldn't find anything useful to work into a lesson, and I just went with Briggs. $\endgroup$ – Richard Mar 14 '15 at 1:14
  • $\begingroup$ I think you did quite well. I surprised myself by finding the method described in matheducators.stackexchange.com/questions/5970/… to use two sheets of ruled paper to construct a rough version of a logarithmic scale, but without sufficient care, it is easy to make mistakes and end up with some serious error. Going through an exercise like mine makes me appreciate your write-up all the more. Gerhard "Appreciates The Precision Of Engineers" Paseman, 2015.03.15 $\endgroup$ – Gerhard Paseman Mar 15 '15 at 20:47
  • $\begingroup$ Here's Numberphile explaining what Napier did: youtube.com/watch?v=vzV50goW_WM $\endgroup$ – john.abraham Jul 19 '15 at 14:32
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Gauss said "You have no idea how much poetry there is in a table of logarithms."

The first paragraph of this paper might get you pointed in the right direction

ON THE DISTRIBUTION OF PRIMES—GAUSS’ TABLES

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    $\begingroup$ I can't find the bit discussing how logs were made. Would you be able to copy the relevant text into your answer? $\endgroup$ – Richard Mar 14 '15 at 1:20

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