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

Update:

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%93Wei...

Update 2:

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%93Wei...

2 Answers

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  • 8 years ago
    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.

  • Anonymous
    8 years ago

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

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