Briggs on Logarithms

Date: 08/28/2001 at 06:29:52
From: Khalid Mahmood
Subject: Logarithms

Dr. Math!        

I want to know how Briggs constructed logarithmic tables of common 
logarithms. The Encyclopedia Britannica says:

"To construct this table Briggs, using about thirty places of 
decimals, extracted the square root of 10 fifty-four times, and thus 
found that the logarithm of 1.0000 0000 000012781 91493 20032 35 was 
0.0000 0000 0000 05551 11512 31257 82702 and that for a number of this 
form (i.e.for numbers beginning with 1 followed by fifteen ciphers, 
and then by seventeen or less numbers of significant figures) the 
logarithms were propotional to these significant figures. He then by 
means of a simple proportion deduced that log (1.0000 0000 0000 1)
= 0.0000 0000 0000 04342 94481 90325 1804, so that, a quantity 1.0000 
0000 0000x(where x consists not more than seventeen figures) having 
been obtained by repeated extraction of the square root of a given 
number, the logarithm of 1.0000 0000 0000x could then be found by 
multiplying x by .0000 0000 0000 o4342..."

Please 

(1) extract the square root of 10 fifty-four times for me

(2) tell me why he took the square root of 10 fifty-four times.

(3) How did he know that the logarithm of 1.0000 0000 0000 
12781 91493 20032 35 was 0.0000 0000 0000 05551 11512 82702  ?

Please explain the paragraph.
                                                                      
I shall be thankful to you for your help.

Date: 08/28/2001 at 12:52:55
From: Doctor Peterson
Subject: Re: Logarithms

Hi, Khalid.

First, let's look at what he was trying to do. Why take the square 
root 54 times?

We know that log(10) = 1. We also know that

    log(a^b) = b log(a)

In particular, therefore,

    log(10^x) = x log(10) = x

When x is an integer, this is simple. But what if we used a very small 
fraction? Then we could find the logarithm of a number very close to 
1. We can raise ten to a small fractional power by taking the square 
root repeatedly:

    log(sqrt(10))             = log(10^0.5)   = 0.5   = 2^-1
    log(sqrt(sqrt(10)))       = log(10^0.25)  = 0.25  = 2^-2
    log(sqrt(sqrt(sqrt(10)))) = log(10^0.125) = 0.125 = 2^-3
    ...

Do you see what I am doing? Each time I take the square root, I am 
raising 10 to a smaller fractional power, and therefore finding a 
number whose log is a smaller number.

If I take the square root 54 times, I am finding

    10^(2^-54) = 1.0000000000000001278191493200323

and I know that its log is

    2^-54 = 5.5511151231257827021181583404e-17
          = 0.000000000000000055511151231257827021181583404

(Of course, I actually found the second number first on my calculator, 
and then raised 10 to that power; Briggs would have actually done the 
square root by hand 54 times, a much longer process. You can take the 
square root repeatedly on a calculator if you want to get the same 
result. I used the calculator on my computer, which gives more 
precision than most.)

You didn't ask me to go on and see how he used this result, but let's 
continue. To find log(1.0000000000000001), we can use the fact that

    log(y) - log(1)    log(x) - log(1)
    --------------- =~ ---------------
         y - 1              x - 1

for x and y close to 1, approximating the log curve by a straight 
line. Then log(y) can be estimated as

    log(y) =~ (y-1)/(x-1) * log(x)

Taking x as the number above, this gives

    log(1.0000000000000001) =~ 
         0.0000000000000001 / 0.0000000000000001278191493200323 * 
         log(1.0000000000000001278191493200323)
       = 0.7823553867474 * 0.00000000000000005551115123125782702
       = 0.0000000000000000434294481903245

That can be used to find other logarithms and build the table.

One more question: why did he use 54? I assume he wanted a particular 
level of precision, so he continued until he got a number close enough 
to 1, and then continued until it looked like 1.000...01..., so that 
he could make an accurate approximation to 1.000...01 by this method.

I checked to see whether we have covered this topic before, and found 
this in our archives (by searching for "briggs log"):

   Logarithms: History and Use
   http://mathforum.org/dr.math/problems/temple.7.12.96.html   

This illustrates the process using simpler numbers.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/