Box 6.3-2: Log10 Tricks (to the 40th degree)


© 1997 & 2003 by Karl Hahn 		KCT logo

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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 logarithms 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 logarithm calculations based solely 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  210 = 1024. And even if you're not, that fact is still true. And 1024 is very near to  1000 = 103. This, together with the rules of logarithms leads to 

   log10(210)  =  10 log10(2) ≈ log10(1000)  = 3
where the symbol, ≈, means "approximately equal to." And dividing 10 out of that you get


   log10(2)  ≈


   3
    
  10
Now indulge me for a minute for preferring to express this fraction as


   log10(2)  ≈


  12
    
  40
My reasons will be evident momentarily.

So what about log10(3)? Well, we have 34 = 81, and that's very nearly 80. So what? you might ask. Well, 80 = 10 × 23, and we already know log10(10) and an approximation to log10(2). So we have 

   log10(34)  =  4 log10(3)  ≈  log10(80) = log10(10) + 3 log10(2)
Using the approximation for log10(2), that becomes


   4 log10(3)  ≈


  40
     + 3 ×
  40

 12
     =
 40

  76
    
  40
Dividing out the 4, we get


   log10(3)  ≈


  19
    
  40
To get an approximation to log10(4), simply observe that 4 is 22. So the log of 4 is twice the log of 2.


   log10(4)  ≈


  24
    
  40
For log10(5), observe that 5 is half of 10. So


   log10(5)  =  log10(10) - log10(2)  ≈


  40
     -
  40

 12
     =
 40

  28
    
  40
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


   log10(6)  ≈


  12
     +
  40

 19
     =
 40

  31
    
  40
Which brings us to 7. Hmmm. Observe that 72 = 49, and that's very nearly 50. Well we already have an approximation for log10(5), so we can find log10(50) simply by adding 1 to it.


   log10(72)  =  2 log10(7)  ≈


  28
     +
  40

 40
     =
 40

  68
    
  40
Divide through by 2 and you've got it.


   log10(7)  ≈


  34
    
  40
Both 8 and 9 are easy.


   log10(8)  =  3 log10(2)  ≈


  36
    
  40
and


   log10(9)  =  2 log10(3)  ≈


  38
    
  40
There you have it -- approximations to the logs base 10 of the numbers from 1 to 10, in 40ths are 

   0, 12, 19, 24, 28, 31, 34, 36, 38, and 40
Hands On Project: Making a Slide Rule
Here's something you can try using some cardboard (file-folder material works too), a compass, and a protractor. This project is thanks to Chuck Fete, who teaches high school math. Cut two circles, one larger than the other. Mount them concentrically with a pin or thumbtack through the common center. Now mark out angles corresponding to the logs base 10 of the numbers, 1 through 9. Since there are 360 degrees in a circle, each fortieth corresponds to 9 degrees. Use the estimates for the logs in fortieths given above to figure out the angle for each number. An image of the final product is shown to the right. By rotating the inner circle, you can multiply by various numbers. The diagram shows three seconds each of multiplying by 1 through 6. Observe how the angles line up between the circles for the different multipliers. See if you can interpret the alignments for yourself based upon the multiplication tables you undoubtedly memorized years ago.

Teachers: For more information on this project

respectively. 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 f(x) = 10 log10(x) in figure 6.3-1 in the main text.

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  10! = 3628800. Our approximation is off by less than 1%.

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 11.2

            table 6.3-2:  40 log10 Approximations
I 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 240 log10(x), and see if you can make sense of the rule of 72, a financial rule of thumb which states that to find out how many years it takes your money to double, divide the annual interest rate (in percentage points) into 72.

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). But then 

   p  =  q log10(n)  =  log10(nq)
Since log10(x) is the inverse function of 10x, it must also be true that 
   10p  =  nq
Remember 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.

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