Mathin’ It Old School

In an earlier post, I proselytized a bit on the importance of logarithms as a core topic not just in mathematics, but in understanding the world. I thought it worth writing on why that discussion is necessary.  It wasn’t so long ago that logarithms motivated themselves because they were essential for common calculations.

The idea is that logarithms turn exponentiation into multiplication via the rule \(\log x^y=y\log x\)  and multiplication into addition via \(\log x y = \log x + \log y\).  (You can’t do anything to \(=\log (x+y)\), however hard my students try.)  You just need to be able to compute logarithms when you are done, but we had handy tables for that.  I have some print books like this one, but you’ll only find them online these days (somewhat ironically).

Let’s see how this works by calculating \(109 \times 31.8\) using logarithms. Books of tables are finite, so we have to represent our numbers as \(A\times 10^B\) where \(1<A<10\). All of our logarithms are base 10, so this guarantees that log A will be between 0 and 1.  So \(109 = 1.09\times 10^2\) and \(31.8 = 3.18\times 10^1\).  If you look at page 12 of the book I posted, you will find this part of a table:

0 1 2 3 4 5 6 7 8 9
10 0.0000 0.0043 0.0086 0.0128 0.0170 0.0212 0.0253 0.0294 0.0334 0.0374
11 0.0414 0.0453 0.0492 0.0531 0.0569 0.0607 0.0645 0.0682 0.0719 0.0755
12 0.0792 0.0828 0.0864 0.0899 0.0934 0.0969 0.1004 0.1038 0.1072 0.1106
13 0.1139 0.1173 0.1206 0.1239 0.1271 0.1303 0.1335 0.1367 0.1399 0.1430
14 0.1461 0.1492 0.1523 0.1553 0.1584 0.1614 0.1644 0.1673 0.1703 0.1732

You read off the logarithm of 1.09 with the first two digits on the left (10) and the last digit along the top (9), and you find log 1.09 is approximately 0.0374.  Another look-up gives log 3.18 is about 0.5024.  Let \(x=109\) and \(y=31.8\), so we want \( xy\). We use \(\log xy = \log x + \log y\), so \(\log 109\times 31.8= \log 109 + \log 31.8\) \(=  \log 1.09 + \log 10^2+ \log 3.18 + \log 10\approx .0374+2+.5024+1=3.5398.\) So 3.5398 is roughly the logarithm of the number we want (or, put another way, we want \(10^{3.5398}\)).  To calculate that, we search our table for the number closest to 0.5398; we’ll recover the 3 later by multiplying by \(1000=10^3\).  We find this on the same page:

0 1 2 3 4 5 6 7 8 9
30 0.4771 0.4786 0.4800 0.4814 0.4829 0.4843 0.4857 0.4871 0.4886 0.4900
31 0.4914 0.4928 0.4942 0.4955 0.4969 0.4983 0.4997 0.5011 0.5024 0.5038
32 0.5051 0.5065 0.5079 0.5092 0.5105 0.5119 0.5132 0.5145 0.5159 0.5172
33 0.5185 0.5198 0.5211 0.5224 0.5237 0.5250 0.5263 0.5276 0.5289 0.5302
34 0.5315 0.5328 0.5340 0.5353 0.5366 0.5378 0.5391 0.5403 0.5416 0.5428

0.5403 is pretty close, so we reverse look-up that entry to find 3.47, giving \(x y\approx 3.47\times 10^3 = 3470\).  If you work it out on your calculator, you get 3466.2.  Not bad — we’ll never get the accuracy of an exact calculation because the table forces some approximation.  That reverse look-up at the end always feels a bit sketchy, but it basically works. Better tables will give better results, of course.

The important thing is that I could multiply those tricky numbers using only addition, a couple of table look-ups, and moving a decimal point around.  Technology has made this technique obsolete. I studied it in high school more than 20 years ago and it seemed quaint then. I’m sure I was among the last cohorts to learn it.

There is also a technique worth mentioning from calculus called logarithmic differentiation that is suffering a similar fate. The idea is that you can calculate the derivative of something like \(f(x) = \frac{x^9 \cos x}{2^x (x^3+9)\ln x}\) more easily by writing \(\ln f(x) = 9\ln x +\ln(\cos x) -x\ln 2-\ln(x^3+9)-\ln\ln x\) and differentiating implicitly, avoiding some horrible product and quotient rule calculations. This was a useful technique during the time when calculators were prevalent but were unable to compute symbolically. But now, if I have a calculation for which logarithmic differentiation would be useful I’m probably going to let Mathematica do it anyway. I no longer teach this technique and I’ve noticed it dropping from modern calculus texts. Computer algebra systems have done to logarithmic differentiation what calculators did to log tables.

Should we mourn these losses to the curriculum?  No.  We do our students — and our discipline — no favors by teaching obsolete skills. But we must account for the loss.  Those log table calculations gave students some practical experience working with logarithms.  We shouldn’t overstate that; log calculation was typically synthesized as a by-the-numbers black box technique.  But at least they saw the box and some students would be willing to think about how it worked.  

Logarithms really are important — they formalize how we think about scale, they give some experience working with transcendental functions, and they are essential in calculus.  We need to make sure we don’t kill the tree when we cut out the dead wood.

Bill Wood is a mathematics professor and tweets as @MathProfBill.  He has just been assigned calculus for the fall semester and is thinking once again about how to get 70 students from 70 mathematics backgrounds on the same page.  That page includes logarithms.

2 thoughts on “Mathin’ It Old School

    1. Bill Wood

      Hi, Mike! Interesting point. I can’t speak much to the accuracy portion, but the speed makes sense. It’s the same as the log table, I guess.. The only real computation is addition, which is easier than multiplication. Looking up logs sucks, but you don’t have to do it much. Same for a computer.

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