Calculating in your head is gradually becoming a lost art, and indeed there is a generation of children coming up these days, most of whom need a calculator to multiply six times nine. But there is nothing like the ability to make rough calculations in your head to both impress your friends and give you a sense of mastery over the numerical world. And as an added bonus, it endows you some measure of protection against those who would dupe you out of your hard-earned money.
As you learned in algebra, the log10 function was once used extensively to carry out multiplication problems. A table of 5-place logrithms was, perhaps, the most used reference book in any math library. But in your head and without a printed log table, you can carry out crude logrithm calculations based soley on a few key multiplication facts. These facts lead you to good working approximations of the logs base 10 of the counting numbers from 1 through 10.
If you are a computer programmer, you probably already know that
log10(210) = 10 log10(2) » log10(1000) = 3where the symbol, », means "approximately equal to." And dividing 10 out of that you get
3 log10(2) »Now indulge me for a minute for preferring to express this fraction as10
12 log10(2) »My reasons will be evident momentarily.40
So what about log10(3)? Well, we have
log10(34) = 4 log10(3) » log10(80) = log10(10) + 3 log10(2)Using the approximation for log10(2), that becomes
40 12 76 4 log10(3) »Dividing out the 4, we get+ 3 ×=40 40 40
19 log10(3) »To get an approximation to log10(4), simply observe that 4 is 22. So the log of 4 is twice the log of 2.40
24 log10(4) »For log10(5), observe that 5 is half of 10. So40
40 12 28 log10(5) = log10(10) - log10(2) »For 6, we know it is the product of 2 and 3. So the approximation to its log will be the sum of the approximations to the logs of 2 and 3. In other words-=40 40 40
12 19 31 log10(6) »Which brings us to 7. Hmmm. Observe that+=40 40 40
28 40 68 log10(72) = 2 log10(7) »Divide through by 2 and you've got it.+=40 40 40
34 log10(7) »Both 8 and 9 are easy.40
36 log10(8) = 3 log10(2) »and40
38 log10(9) = 2 log10(3) »There you have it -- approximations to the logs base 10 of the numbers from 1 to 10, in 40ths are40
0, 12, 19, 24, 28, 31, 34, 36, 38, and 40respectively. If you don't believe me, divide each of them by 4 on a calculator and see how close they come to values listed as
Now let's do a quickie calculation using these. Suppose you wanted to
know 10!. That's the product of the numbers from 1
to 10. And we know that the log of the product is the sum
of the logs.
So add these numbers up (which is an exercise many
folks can do in their heads) and you get 262. When you divide
by 40, you get 6 with a remainder of 22.
The 6 tells us that 10! is somewhere between
one million and ten million (that is, it's in the 106
decade, and 106 will be the characteristic).
Now what about the remainder? It tells us the mantissa. Find
where the remainder fits in the list of numbers above. It falls three
fifths of the way between 19 and 24, which are the
approximations for log10(3) and log10(4)
respectively. Three fifths of the way between 3 and 4
is 3.6. So the answer is 3.6 million. How accurate
is that? Well, in reality
Here's another quickie table that's useful for doing still more calculations.
x 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 40 log10(x) 0.17 0.34 0.51 0.68 0.85 1.0 1.2 1.3 1.5 x 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 40 log10(x) 1.7 3.2 4.6 5.8 7.0 8.2 9.2 10.2 11.1 table 6.3-2: 40 log10 ApproximationsI had to cheat and use a calculator to make this, so for it to be useful, you either have to memorize it or make a cheater-card and keep it in your wallet. Then, for example, if somebody wanted to know what $1000 would be left for 25 years at 6% interest, you could take the approximation for log10(1.06), multiply it by 25 to get 25, then remember that's in 40ths and that 25/40 is one fourth of the way between the approximation we had for log10(4) and log10(5). Then you'd know that you would end up with about $4250 (the actual is $4292).
With all this in mind, try multiplying all the approximations by 6,
so that each represents
And by the way, how do you think I knew that the logs base 10 of
the integers 2 through 9 are irrational? Pick any natural
number that is not a power of 10. Call it n. If
log10(n) were rational, then there would be two
natural numbers, p and q, such that
p = q log10(n) = log10(nq)Since log10(x) is the inverse function of 10x, it must also be true that
10p = nqRemember that n, p, and q are all counting numbers. See if you can make an argument based upon prime factorization of the two sides of the above equation to show why this equation is impossible unless n is a power of 10.
email me at hahn@netsrq.com