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
_____________________________________________

Estimating Logarithms without Using a Calculator

Date: 07/28/2005 at 13:02:30
From: Amanda
Subject: log problem

I am trying to find the following answer without a calculator:

log_4 12

From reading respones from Dr. Math, I believe the statement to read: 
what is the power(exponent) that raises a base of 4 to make 12.  I can 
come to a broad conclusion that the answer is between 1 and 2.  Other
than that I have the problem 4^x = 12.  How do I solve?

Thanks


Date: 07/28/2005 at 14:53:48
From: Doctor Douglas
Subject: Re: log problem

Hi Amanda.

Excellent work so far.  Indeed you are solving 4^x = 12 for x, and it 
is clear that 1 is too small and 2 is too big.  A good way to continue 
proceeding is to factorize 12:

  12 = 4 * 3.

So we can write

  4^x = 4 * 3

Taking the log of each side, we get

  x*log(4) = log(4*3) = log(4) + log(3)

and we can solve for x as

  x = [log(4) + log(3)]/log(4)
     = 1 + log(3)/log(4)

which is about as simple as one can get without resorting to a 
calculator to compute the logarithms.  Now it gets a little more 
complicated to proceed with finding a numerical answer if we are not 
permitted the use of a calculator.

Here's a way to do so, where I assume that the logs are taken in base 
10:

   log(3):   Consider powers of 3.  3^2 = 9, which is close to 10^1.
             Hence 2*log(3) is approximately equal to 1*log(10) = 1,
             or log(3) :=: 1/2, where I use the colons to indicate
             "approximately equal to".  For more accuracy, we can
             take higher powers of 3 until we find another one that
             is close to a power of 10:  3^21 :=: 10^10, so that
             log(3) = 10/21 = 0.47619.  

   log(4):   We do the same with powers of 4:  4^5 :=: 10^3, so 
             log(4) :=: 3/5 = 0.6, to pretty good accuracy.  I say
             pretty good accuracy because 4^5 = 1024 and 10^3 = 1000
             only differ by 2.4%.  In fact we can't do much better
             until going all the way to 4^98 :=: 10^59, whereupon
             log(4) :=: 59/98 = 0.6020408...
             
The calculations above require nothing more than simple arithmetic and 
a knowledge of the rules of logarithms, although I admit that the
multiplication can be painstaking.

So our estimates are

  x = 1 + log(3)/log(4)
    
   :=: 1 + (1/2)/(3/5) = 1.83...          by computing 3^2 and 4^5

   :=: 1 + (10/21)/(59/98) = 1.791...     by computing 3^21 and 4^98

Using a calculator, I obtain log(3) = 0.477121255.... and log(4) =
0.602059991..., and a final answer of x = 1.792481250360578...  This 
shows that our estimates aren't too bad.

Remark:  we could have just started from powers of 12 in the 
  beginning:  12^5 = 248832 :=: 262144 = 4^9, so an estimate for
  our answer is x = log(12)/log(4) = 9/5 = 1.8, which is not too
  bad.  However, if we do it this way then we have to be able to
  recognize when a power of 12 is in fact close to a power of 4.
  It's a lot easier to do these computations by using base 10 as an
  intermediary, since it is very easy to recognize the powers of 10.
  To do this efficiently, it was helpful in the beginning to factor
  the original 12 into smaller numbers to reduce the amount of 
  multiplicative arithmetic.

- Doctor Douglas, The Math Forum
  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/