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How do we know if Log(2)_3 is not equal to something like ((x^y)+a)

 
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Aug11-12, 06:50 AM   #1
 

How do we know if Log(2)_3 is not equal to something like ((x^y)+a)

How do we know if Log(2)_3 is not equal to something like ((x^y)+a) ,for rational a,x,y ?
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Aug11-12, 07:16 AM   #2
 
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Use this: http://en.wikipedia.org/wiki/Gelfond...neider_theorem
Aug11-12, 07:45 AM   #3
 
thanks, but how is
"If α and β are algebraic numbers with α ≠ 0,1 and if β is not a rational number, then any value of αβ = exp(β log α) is a transcendental number."
equivalent to
"if α and γ are nonzero algebraic numbers, and we take any non-zero logarithm of α, then (log γ)/(log α) is either rational or transcendental"
?
Aug11-12, 07:53 AM   #4
 
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How do we know if Log(2)_3 is not equal to something like ((x^y)+a)

What is
[tex]\alpha^{\log(\gamma)/\log(\alpha)}[/tex]
?
Aug11-12, 08:05 AM   #5
 
Quote by micromass View Post
What is
[tex]\alpha^{\log(\gamma)/\log(\alpha)}[/tex]
?
you mean [tex]\gamma [/tex] ?
Aug11-12, 08:06 AM   #6
 
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Quote by limitkiller View Post
you mean [tex]\gamma [/tex] ?
Yes.

So, IF [itex]\log(\gamma)/\log(\alpha)[/itex] were an algebraic nonrational number, then by applying Gelfond-Schneider we get ...
Aug11-12, 08:10 AM   #7
 
I get it
Aug11-12, 08:10 AM   #8
 
thanks
Aug11-12, 08:11 AM   #9
 
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Of course, to be able to apply Gelfond-Schneider to [itex]\log(2)/\log(3)[/itex], we must first prove that it's not rational...
Aug11-12, 08:22 AM   #10
 
Which is easy, isnt it?

since 2 ^a for integer a is never a power of 3...
Right?
Aug11-12, 08:24 AM   #11
 
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Quote by limitkiller View Post
Which is easy, isnt it?

since 2 ^a for integer a is never a power of 3...
Right?
Yeah, that's it (except for a=0 of course, but that's also not possible). I just wanted to point it out in case you missed it
Aug11-12, 08:27 AM   #12
 
Quote by micromass View Post
I just wanted to point it out in case you missed it
I actually did.
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