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

Question

I'm self-studying and a beginner in proofs and here's my take.

Let x be the logarithm of 7 to base n where n is an integer. Then $7^x = n$. If n is an integer, then x should be any nonnegative integer, that is $ x \geq 0$. Thus $\log_7 (n)$ is an integer.

Is this enough for the integer part? I'm certain that this does not answer the irrational part. Also when it says on the question '..either an integer or irrational', does the 'integer' mean that it includes negative integers?

Thanks a lot!

Answer 1

We may in fact show that $\log_p(n)$ is either integer or non-rational for any prime $p$:

Suppose $\log_p(n)$ is a non-integer rational; then

$\log_p(n) \in \Bbb Q \setminus \Bbb Z; \tag 1$

thus,

$\exists r, s \in \Bbb Z, \; \log_p(n) = \dfrac{r}{s}, \tag 2$

and we may of course take

$\gcd(r, s) = 1; \tag 3$

note that these stipulations force

$s \ne 1, \tag 4$

lest (1) fail to bind; now (2) implies

$n = p^{r/s}, \tag 5$

from which

$n^s = p^r; \tag 6$

thus $p$ is the only prime factor of $n$, and so

$\exists m \in \Bbb N, n = p^m; \tag 7$

then we have, in light of (4),

$p^r = n^s = (p^m)^s = p^{ms}$ $\Longrightarrow r = ms \Longrightarrow s \mid r \Rightarrow \Leftarrow \gcd(r, s) = 1, \tag 8$

a contradiction which precludes (1). $OE\Delta$.

In closing, I would like to say that I believe our OP user587266's assertion that "If n is an integer, then x should be any nonnegative integer, that is $ x \geq 0$" is in and of itself insufficient; indeed, that is one of the things we are trying to prove. So, respectfully, no, that is "not enough for the integer part." Finally, we observe that though $\log_p(n)$ may indeed be an integer, it cannot be a negative integer for $n \in \Bbb N$; for if

$\log_p(n) = -m, \; m \in \Bbb N, \tag 9$,

then of course

$n = p^{-m} = \dfrac{1}{p^m} < 1; \tag{10}$

but the contradicts

$n \in \Bbb N. \tag{11}$