Logarithms Lecture 7: Version 2: History

In this lecture, I want to present a different motivation and way of thinking about logarithms. This is the historically the way they were discovered and first used, so I will say a bit about the history as well. The central fact is still the conversion of multiplication to addition:

log (a x b) = log(a) + log(b)

for any non-zero numbers a and b.

Logarithms were developed independently by the Scottish theologian and mathematician John Napier, and by the Swiss clock maker Joost Bürgi, in the late 1500s and early 1600s. Napier was more successful in publicizing his method, so his name is usually attached to the discovery.

Both of them had the same goal: to simplify numerical computation. Multiplication, division, powers, and roots are all very tough to do by hand. Each of them cooked up a system which had the central property above: they each found a way of making a function (call it f) such that, for any two non-zero numbers a and b,

f(a x b) = f(a) + f(b)

We will see that any function with this property is essentially a log to some base; this is a defining property of logs. Napier’s version was, up to a scaling, a log base e; later, the English mathematician worked with Napier to produce a version of log base 10.

Napier didn’t define logs the way we did; he had a very curious system of two moving particles. It’s too far out of the way for me to explain here. If you’d like to know how it worked, and would like more of the history, you can start with Logarithms: The Early History of a Familiar Function, by Kathleen Clark and Clemency Montelle. (The linked article is one chapter of a series; see the links at bottom if you want even more information.)

Napier’s funny-looking definition was just made to obey the central property, of converting multiplication to addition. This property would allow one to do multiplication much more easily, and to do division, powers, and roots as well.

Multiplication using logs

We’ve already seen examples of doing multiplication, for example, when I showed you how to compute the number of possible Vermont license plates, or when I asked you to compute the number of seconds in a year. Let’s do another example to remind you how it works. Suppose I want to multiply

32,768 x 512.

The number 32,768 is about 4.52 zeroes, 512 is about 2.71 zeroes, so their product is about 7.23 zeroes. Then, 7.23 zeroes is about 17,000,000. Written more mathematically,

$\log(32,768 \times 512) = \log(32,768) + \log(512)\simeq 4.52 + 2.71=7.23$ ,

so the product we are looking for has a log of about 7.23; then we look for a number with that log:

$\log(17,000,000)\simeq 7.23$ .

The answer to 32,768 x 512 should be about 17,000,000. Actual answer:

(How did I get those extra decimals? We know that approximately, log(3) = 0.5, and log(4) = 0.6. The number 3.2768 rounds to 3.3, which is about 30 per cent of the way between 3 and 4, so the log should be 30 percent of the way between log(3) and log(4), i.e. 0.53. That’s how I got log(32,768)=4.53. I got the other second decimal places similarly. Try it!)

This is great for mental computation. If you want a more accurate answer, you can work out a set of tables of logarithms; that is what Napier, Bürgi, and Briggs did.

A typical log table, a version of which used to appear in the back of most math textbooks. (Click to get a bigger, more legible version.)

So you can look up logs on a table, and find, to four decimal places,

$\log(32,768 \times 512) = \log(32,768) + \log(512)\simeq 4.5154 + 2.7093=7.2247$ .

Now you can use that same log table, but read it backwards, to find that a number with a log of 7.2247 is 16,776,000. Look at the answer again–we’ve gotten very close!

In fact, there is no need to tabulate that many logarithms; we just need to tabulate the logarithms between 1 and 10. This is because of what we’ve seen earlier: e.g.

$\log(32,768) = \log(3.2768\times 10^4) = 4 + \log(3.2768)$ ;

in this way, we only have to look up log(3.2768), which is between 1 and 10.

Of course, this is largely obsolete today, now that we have electronic calculators. But before calculators, this was an enormous savings of effort: this process is immensely easier than multiplying by hand. Compare the work of multiplying 32,768 x 512 longhand, to just adding the numbers 4.5154 + 2.7093. Laplace is said to have remarked, “by cutting short their labors, logarithms have doubled the life of the astronomer”.

From the early 1600s up to the 1970s, every student of mathematics, every mathematician or engineer or physicist, would keep a set of log tables, and use it to perform multiplications in this way. The slide rule was also a calculating instrument also based on logarithms.

A slide rule. See this site by Peter Alfeld for image credits, and more information on slide rules if you are interested!

Problem 1: For practice, do a few more multiplication questions. That is, use logarithms to do some approximate multiplications mentally. You can of course use very large numbers, that makes it no harder. Check your answers on a calculator and see how close you get. If you are feeling ambitious, try to estimate second decimal places, as I did in the example above, to get more accurate answers. With some practice, you can do some impressive multiplications quite quickly, at least approximately. You can impress your friends, depending on the sort of friends you have.

Division using logarithms

It isn’t just multiplication! The introduction of logarithms made many computations vastly easier. The next thing we can do is division.

Note that longhand division is a very tedious process, particularly if the denominator is large. Finding a number like

$\dfrac{16,777,216}{32,768}$

is quite unpleasant by hand. (Some American schools have even stopped teaching how to do this by hand, (“long division”), so you might not know the method!) But logarithms make it much easier.

The fundamental property of logarithms implies another property: for any two non-zero numbers a and b,

$\log\left(\dfrac{a}{b}\right) = \log(a) - \log(b)$ .

This makes a kind of sense, comparing to the multiplication rule, right? But we can do better, and actually prove it from the multiplication rule:

Problem 2: Use the fact that $\dfrac{a}{b}\times b=a$ to prove the division rule for logarithms shown above.

This makes division almost as easily as multiplication. Roughly, mentally:

$\log\left(\dfrac{16,777,216}{32,768}\right)=\log(16,777,216)-\log(32,768)$
$\simeq 7.23 - 4.52 = 2.71$ ,

so the log of the answer we are looking for is 2.71. A number with a log of 2.71 is about 510. (We know from above that the answer is 512!)

If we needed a more accurate answer, we could refer to log tables, as described above. So, to divide, all we need is a log table, and to be able to subtract!

Problem 3: Make up a few examples of divisions, and estimate them mentally by using logarithms. Again, the numbers can be large. Check your answers on a calculator.

Powers and roots

Logarithms make other operations easier as well. To show these examples, we will need one more fundamental property of logarithms.

For an example, remember that the number 2 represents about 0.3 of a zero. Expressed more formally, log(2) = 0.3 approximately. Now, what happens if we have a power of 2? For example, what is $\log(2^{13})$ ? Well,

$2^{13}=2\times 2 \times 2 \times \dotsb \times 2$ (13 times),

so if each two is worth 0.3 zeroes, then 13 twos must be worth 13 x 0.3 = 3.9 zeroes:

$\log(2^{13})=13\times\log(2)$ .

This same reasoning would work with any whole number power:

$\log(2^x)=x\times\log(2)$ ,

and there is nothing special about 2, it would work with any non-zero number in place of 2:

$\log(a^x)=x\times\log(a)$ .

I’ve only given the reasoning for positive whole number powers, but in fact it works for negative or fractional numbers too: for any non-zero number a, and any number x,

$\log(a^x)=x\times\log(a)$ .

It is also possible to prove this property from the fundamental multiplication property. We could do it with what we know, but it’s a bit long, so I’m going to skip it. (You can take that as a challenge problem if you’d like.)

This means that I can use logs to compute powers of numbers. I almost did it above: if I want to quickly mentally calculate $2^{13}$ , I could observe

$\log\left(2^{13}\right)=13\times\log(2)\simeq 13\times 0.3 =3.9$ ,

and note that $\log(8,000)\simeq 3.9$ , so $2^{13}\simeq 8,000$ . (The exact answer is 8,192. If we had log tables and used pencil and paper, we could get a lot closer.)

This works even for forbiddingly large numbers. (Very large numbers are important in some problems that have to do with counting.) For example:

Problem 4: Estimate $2^{1000}$ .

Did you get an answer? Not too hard, right? Let’s check it on the calculator:

Too big for the calculator to handle!

I can also compute roots. From what we’ve reasoned above, a square root is a power of 0.5. (Remember, for example, we reasoned that a “half a zero” should be $\sqrt{10}$ , in other words, $10^{0.5}=\sqrt{10}$ .) So let’s say we want to compute

$\sqrt{52,000}$ .

We can estimate $\log{52,000}\simeq 4.72$ . So

$\log(\sqrt{52,000})=\log\left(52,000^{0.5}\right)=0.5\times\log(52,000)$
$\simeq 0.5\times 4.72 = 2.36$ .

The log of our answer is about 2.36. But $\log(230)\simeq 2.36$ . So

$\sqrt{52,000}\simeq 230$ .

Calculator says:

Again, for us it is a neat party trick. But before there were calculators, doing a square root by hand was misery without log tables. With log tables, all you had to do was divide a number by 2.

Problem 5: Estimate the following mentally. Then check on a calculator.
a. $3^{27}$
b. $320^{17}$
c. $(0.5)^{10}$

Problem 6: Estimate two or three square roots mentally, using logs. Check your answers on a calculator.

Multiplication using logs

Division using logarithms

Powers and roots

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