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If I have two integers $a,b > 1$. Is

$\ln(a) - \ln(b)$

always either irrational or $0$. I know both $\ln(a)$ and $\ln(b)$ are irrational.

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up vote 28 down vote accepted

If $\log(a)-\log(b)$ is rational, then $\log(a)-\log(b)=p/q$ for some integers $p$ and $q$, hence $\mathrm e^p=r$ where $r=(a/b)^q$ is rational. If $p\ne0$, then $\mathrm e=r^{1/p}$ is algebraic since $\mathrm e$ solves $x^p-r=0$. This is absurd hence $p=0$, and $a=b$.

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Wow, that was really simple, I feel kind of silly not seeing that myself. Thanks. – Zert Oct 30 '12 at 10:14

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