Converting a Base 2 Log

Date: 01/31/2002 at 23:46:20
From: Sue
Subject: Powers

I think the right equation is 2147483647 = 2^(x) - 1, but I don't know 
how to figure it out. Please help if you can.

Thank you very much.

Date: 02/01/2002 at 01:20:53
From: Doctor Jeremiah
Subject: Re: Powers

Hi Sue,

Your equation  2147483647 = 2^(x) - 1  is exactly right.

                   2147483647 = 2^(x) - 1
               2147483647 + 1 = 2^(x) - 1 + 1
                   2147483648 = 2^(x)

To figure it ouw you have to either do it by trial and error or use 
logarithms.  A logarithm gives you the exponent. For 10^x you would 
use a base 10 log ( which you can write as log() ) and for e^x you 
would use the natural log ( which you can write as ln() ). And for 2^x 
you need to use the base 2 log which I will write as lg2() :

                   2147483647 = 2^(x) - 1
               2147483647 + 1 = 2^(x) - 1 + 1
                   2147483648 = 2^(x)
            lg2( 2147483648 ) = lg2( 2^(x) )

Now, log( 10^x ) = x and ln( e^x) = x so lg2( 2^x ) = x

                   2147483647 = 2^(x) - 1
               2147483647 + 1 = 2^(x) - 1 + 1
                   2147483648 = 2^(x)
            lg2( 2147483648 ) = x

How do you calculate a base 2 log? Well, most calculators have a 
natural log and a base 10 log, but a base 2 log is not there.

But you can convert a base 2 log into some other log that you do have 
the ability to calculate.

  log base a ( x ) / log base a ( b ) = log base b ( x )

So let's convert our base 2 log to a base 10 log

                   2147483647 = 2^(x) - 1
               2147483647 + 1 = 2^(x) - 1 + 1
                   2147483648 = 2^(x)
            lg2( 2147483648 ) = x
 log( 2147483648 ) / log( 2 ) = x

We can calculate base 10 logs, so now we can figure out what x is.  
Please write back if you want to talk about this more.

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