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Re: prove or disprove: there are no whole numbers which work in the equation log(base2)3=p/q (90 Views)
Posted by: joejoe (IP Logged)
Date: September 08, 2009 03:29PM
Bear in mind I'm not quite sure of the answer but I say "No" there are no intergers p/q which satisfy the equation log(base2)of3=p/q
log(base2)of3
USE THE FORMULA log(baseA)ofB=log(baseC)ofB/log(baseC)ofA
log(base2)of3=log(Base10)of3/log(base10)of2
SO
log(Base10)of3/log(base10)of2=p/q
REMEMEBR THE DEFINITION, AN INTERGER IS A WHOLE NUMBER.
log(base10)of3=p, 10^p=3 (p is some number between 0~1)
log(base10)of2=q, 10^q=2 (q is some number between 0~1)
Q is not an interger. P is not an interger
log(base2)of3
USE THE FORMULA log(baseA)ofB=log(baseC)ofB/log(baseC)ofA
log(base2)of3=log(Base10)of3/log(base10)of2
SO
log(Base10)of3/log(base10)of2=p/q
REMEMEBR THE DEFINITION, AN INTERGER IS A WHOLE NUMBER.
log(base10)of3=p, 10^p=3 (p is some number between 0~1)
log(base10)of2=q, 10^q=2 (q is some number between 0~1)
Q is not an interger. P is not an interger
| Subject | Written By | Posted |
|---|---|---|
 prove or disprove: there are no whole numbers which work in the equation log(base2)3=p/q (156 Views)
|
5sprogs | 08/30/2009 05:22PM |
| Re: prove or disprove: there are no whole numbers which work in the equation log(base2)3=p/q (90 Views) |
joejoe | 09/08/2009 03:29PM |
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