Are there any rational numbers x such that ln(x) is algebraic (log base e)?

2013-05-20T08:26:42Z

Yes - I'm looking for non-trivial solutions (values other than x=1). It looks like it can be proven that no such numbers exist based on the Lindemann-Weierstrass Theorem:

http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

2013-05-20T08:26:47Z

Yes - I'm looking for non-trivial solutions (values other than x=1). It looks like it can be proven that no such numbers exist based on the Lindemann-Weierstrass Theorem:

http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

?2013-05-21T20:58:59Z

Favorite Answer

Indeed Lindemann's piece of Lindemann-Weierstrass is enough: it says e^a is transcendental for every non-zero algebraic number a. If there were a rational x>0, x != 1 such that ln(x) = a != 0 were algebraic, then x = e^a would be transcendental and rational, a contradiction.

10

Anonymous2013-05-20T03:09:06Z

ln(1)=0 satisfies that. Are you looking for a different number?

00

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