Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

How Does a Calculator Find Log?


Date: 03/23/98 at 11:43:21
From: CMLee
Subject: How does a calculator find log?

My teacher has told me that there is no specific formula to derive 
logarithms in decimal form. I was wondering if this was true. 

If it is, then how does a calculator or computer perform calculations 
to find approximations of particular logs, say log5. Also where are 
the tables of logarithms derived from? 


Date: 03/23/98 at 13:07:05
From: Doctor Rob
Subject: Re: How does a calculator find log?

There are infinite sums which give the values of logarithms.  
Examples are the following:

   log_e(1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - x^6/6 + ... 
   (-1 < x < 1)

   log_e([n+1]/[n-1]) = 2/n + 2/(3*n^3) + 2/(5*n^5) + 2/(7*n^7) + ...

In particular, to find the log of 5 base e, you can use the first 
equation with x = -4/5 to compute the log of 1/5, and the log of 5 
is the negative of that answer.  

The terms in these sums get very small after a while, particularly if 
the value of x is small, or the value of n is large. To approximate a 
logarithm to a certain accuracy, you only need to compute a finite 
number of these terms. The smaller x is, the fewer terms are needed.  
To compute log_e(2) to four decimal places of accuracy, put n = 3 in 
the second sum, and use the first four terms. The fifth term is about 
.00001129, the sixth term .000001026, and subsequent terms are even 
smaller, so ignoring them does not affect the approximate answer. 
Once you know log_e(2), you can get log_e(5) by using the first sum 
with x = 1/4 and 11 terms, to get log_e(5/4), then add twice log_e(2).  
Then log_e(5) + log_e(2) = log_e(10). This is the conversion factor 
you need to convert to log_10(x):

   log_10(x) = log_e(x)/log_e(10).

This is how the logarithm tables are computed, and basically how 
electronic calculators compute, too.  

One strategy is to have precomputed and stored log_e(2) and log_e(10).  
Divide your number by 2 until it is between 2/3 and 4/3, say k 
divisions are required.  Alternatively, multiply your number by 2 
until it is between 2/3 and 4/3, say (-k) multiplications are needed.
Use the first sum above with x equal to that result minus 1 (which is
between -1/3 and 1/3), then add k*log_e(2) to that answer. Finally, 
divide by log_e(10).

Another way to approximate the logarithm of a given number is to use 
the square root function and the definition of log_10(x) as the 
exponent to which 10 must be raised to get x.  As an example let's 
take log_10(5).  

   We know log_10(10) = 1, which is too large, and log_10(1) = 0,   
   which is too small.  log_10(sqrt[10]) = 1/2, and we can compute

   10^(1/2) = sqrt(1*10) = 3.16227766..., too small.

   Then since sqrt(10) < 5 < 10, we know that 1/2 < log_10(5) < 1.  
   Now compute: 

   10^(3/4) = sqrt[10*sqrt(10)] = 5.52341325... > 5, too big.

   This implies that 1/2 < log_10(5) < 3/4.  We started with an 
   interval of length 1. Each square root computation has cut the 
   length of the interval in which we know log_10(5) lies in half.  
   After two square roots, we have an interval of length 1/4.  
   Eventually we will have an interval which is smaller than 1/10000 
   (at least 14 square roots are required), and that will tell us 
   log_10(5) to four decimal places:

   10^0             = 1,             too small,
   10^1             = 10,            too big,
   10^(1/2)         = 3.16227766..., too small,
   10^(3/4)         = 5.52341325..., too big,
   10^(5/8)         = 4.21696503..., too small,
   10^(11/16)       = 4.86967525..., too small,
   10^(23/32)       = 5.23299114..., too big,
   10^(45/64)       = 5.04806571..., too big,
   10^(89/128)      = 4.95806824..., too small,
   10^(179/256)     = 5.00286461..., too big,
   10^(357/512)     = 4.98041606..., too small,
   10^(715/1024)    = 4.99162771..., too small,
   10^(1431/2048)   = 4.99724300..., too small,
   10^(2863/4096)   = 5.00005301..., too big,
   10^(5725/8192)   = 4.99864781..., too small,
   10^(11451/16384) = 4.9993503...,  too small,
   10^(22903/32768) = 4.9997016...,  too small,
   10^(45807/65536) = 4.9998773...,  too small,

  So 45807/65536 = 0.6989593 < log_10(5) < 0.6989746 = 2863/4096.   
  Thus, to five decimal place, log_10(5) is either 0.69896 or 0.69897, 
  and to four decimal places, log_10(5) = 0.6990, which your   
  calculator or tables can verify.  

The pattern in the fractions above is that the denominator doubles 
at each step, and the numerator is double plus 1 if the previous 
answer was too small, and double minus 1 if the previous answer 
was too big. Each power of 10 is computed as the square root of the 
product of the last "too big" power times the last "too small" power.

I hope this answers your question.

-Doctor Rob,  The Math Forum
Check out our Web site http://mathforum.org/dr.math/   
    
Associated Topics:
High School Logs

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/