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Converting a Base 2 Log

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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

Thank you very much.
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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.

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

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