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### 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..."

(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  ?

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/
```
Associated Topics:
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