Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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 '14 at 7:17

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.

share|cite|improve this answer
n must be a power of 7, can you prove this also ? – mehdi Aug 5 '12 at 18:11
@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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.