Long Numbers and Logarithms

Date: 10/11/2002 at 20:05:43
From: Richard
Subject: Numbers, exponents

Hi, 

I have a problem figuring out how many terms are there when I raise a 
number to another large number or a large exponent. Is there a formula 
for these problems?

Example:
         
   (12)^5 = 248832    There are 6 terms in the answer.  
   (12)^6 = 2985984   There are 7 terms in the answer.  
   (12)^7 = 35831808  There are 8 terms in the answer.

   (12)^(214)         How many terms will there be?

   How about (123)^214?  How many terms are there?

Date: 10/12/2002 at 12:50:26
From: Doctor Ian
Subject: Re: Numbers, exponents

Hi Richard,

Usually if you want to work with large numbers of digits, you have to 
use logarithms. 

For example, consider

       12^5 = 12 * 12 * 12 * 12 * 12

  log(12^5) = log(12 * 12 * 12 * 12 * 12)

            = log(12) + log(12) + log(12) + log(12) + log(12)

            = 5 * log(12)

            = 5 * 1.08

            = 5.40

So you'd expect to see 6 digits. Now, suppose we look at a higher 
exponent, like 20.

  log(12^20) = 20 * log(12)

             = 20 * 1.08

             = 21.6

So here, we'd expect to see 22 digits, not 21. And

  log(12^214) = 214 * log(12)

              = 214 * 1.08

              = 231.12 

which would have 232 digits.  Because this is close to 231, it's a good idea to use more digits in the approximation of the logarithm, to make sure we're not over-estimating:

  log(12^214) = 214 * log(12)

              = 214 * 1.07918

              = 230.94452

So we'd really expect to see 231 digits.   

Does this help?

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