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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 ?

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See this as well. –  J. M. Aug 5 '12 at 23:54
    
As stated your statement is not true. For instance if $n=\sqrt{7}$, then $\log_7\sqrt{7}=\frac{1}{2}$. I think that you need $n$ to be a non-zero natural number. –  Baby Dragon Jan 2 at 7:17
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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.

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n must be a power of 7, can you prove this also ? –  mehdi Aug 5 '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 5 '12 at 18:12
    
Can you help to explain where exactly the contradiction is and what a = kb is contradicting? –  user44045 Oct 8 '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 8 '12 at 23:41
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