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Show that $\log_7 n$ is either an integer or an irrational number where n is a positive number.

I assumed that it is rational and tried to get a contradiction for $\log_7 n = a/b$, where b does not divide a, but how can I show that $7^{a/b}$ is not an integer to achieve a contradiction since n is an integer ? If I can exclude rational numbers from the range of log function then it is either integer or irrational.

Or do you suggest other methods ?

mehdi
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1 Answers1

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Interesting; usually one would assume not just that $b$ doesn't divide $a$ but that $a$ and $b$ are coprime, but in this case your assumption that $b$ doesn't divide $a$ is enough.

If $7^{a/b}=n$, then $7^a=n^b$. Thus $n$ must be a power of $7$, so we can write $n=7^k$ and thus $7^a=7^{kb}$, so $a=kb$, contradicting the assumption.

joriki
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  • n must be a power of 7, can you prove this also ? – mehdi Aug 05 '12 at 18:11
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    @mehdi: It follows from the fundamental theorem of arithmetic. Since the only prime in the prime factorization of $7^a$ is $7$ and prime factorizations are unique, $n^b$, and therefore $n$, can't contain any other primes. – joriki Aug 05 '12 at 18:12
  • Can you help to explain where exactly the contradiction is and what a = kb is contradicting? –  Oct 08 '12 at 20:56
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    @acw622: It contradicts the assumption that $b$ doesn't divide $a$. By definition $b$ divides $a$ iff there is an integer $k$ such that $a=kb$. – joriki Oct 08 '12 at 23:41