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Briggs Logarithms


Date: 09/06/2001 at 23:01:40
From: Khalid Mahmood
Subject: Briggs Logarithms

Dr. Math!

I have a difficulty with the following paragraph from the Encyclopedia 
Britannica, page 876:

"To find the logarithm of 2, Briggs raised it to the tenth power, viz. 
1024, and extracted the square root of 1.024 forty seven times, the 
result being 1.00000 00000, 00000, 16851 60570 5394977. Multiplying 
the significant figures by 4342... he obtained the logarithm of this 
quantity, viz. 0.00000, 00000, 00000, 07316 55936 90623 9336, which 
multiplied by 2^47 gave 0.01029 99566 39811 95265 277444, the 
logarithm of 1.024, true to 17 or 18 places. Adding the characteristic 
3, and dividing by 10, he found (since 2 is the tenth root of 1024) 
log2 = . 30102 99956 63981 195. Briggs calculated in a similar manner 
log 6 and then deduced log 3."

I have the following questions

1. Why did Briggs raise 2 to the tenth power- why not to the ninth or 
   eleventh power?
2. Why did he extract the square root of 1.024 instead of 1024?
3. Why did he extract the square root of 1.024 forty seven times? 
4. Only by multiplying by 4342 how can he get the logarithm?
5. log (1.00000 00000 00000 16851 60570 5394977) = 
   (0.0 .000000000 00000 07316 55936 90623 9336)  
   Multiplying both sides by 2^47 we get
   log 1.024 = 0.01029 99566 39811 95265 277444
6. Adding the characteristic 3 and dividing by 10 he found 
   log 2 = . 30102 99956 63981 195 . 
                                                                      
Please explain.


Date: 09/07/2001 at 14:36:46
From: Doctor Peterson
Subject: Re: Briggs Logarithms

Hi Khalid.

The answer to the first two questions is easy: Briggs is looking for 
numbers close to 1, so he takes a power of 2 that is close to a power 
of ten, and divides by that power of ten to get a number close to 1.

Now, why 47 square roots? Again, he is looking for numbers very close 
to 1, so he raises 1.024 to a small power (taking the square root 47 
times raises it to the 2^-47 power); I can only guess why he chose 47, 
but it looks as if he wanted to get the number small enough so that, 
with the desired level of precision, he had only 17 significant digits 
in the number, corresponding to the number of significant digits in 
his value for log (1.000 000 000 000 1). His goal is to get a certain 
number of significant digits in each number he obtains, rather than 
just calculate digits without knowing whether they are accurate.

Again, why multiply by 0.4342...? Recall that

log(1.000 000 000 000 1) = 0.000 000 000 000 043 429 448 190 325 180 4

and that, when x and y are close to 1,

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

Applying this to x = 1.000 000 000 000 1, we get

    log(y) = (y-1) * 0.000 000 000 000 043 429 / 0.000 000 000 000 1
           = (y-1) * 0.43429...

Having a value for log(1.024^(2^-47)), we see that this is equal to

    2^-47 * log(1.024)

Multiplying this by 2^47 therefore gives us

    log(1.024) = 0.010 299 956 639 811 952 652 774 44

But this is

    log(10^-3 * 2^10) = -3 log(10) + 10 log(2) = -3 + 10 log(2)

and we can solve

    10 log(2) - 3 = x

to find

    log(2) = (x + 3) / 10

So adding 3 and dividing by 10 gives the log of 2.

Putting this all together, Briggs used this formula to find log(2):

log(2) =
[2^47 * log(1 + 10^-16)*10^16 * ((10^-3 * 2^10)^(2^-47) - 1) + 3] / 10
        \___________________/    \____________/
             0.43429...              1.024

Briggs did some pretty good thinking, didn't he? Modern computers, 
when programmed intelligently, use similar tricks to make the best use 
of their capabilities to hold small numbers with great precision.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Logs

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