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The following answer provides a proof that $\log_7{n}$ is either an integer or irrational. https://math.stackexchange.com/a/179198

In a comment, the author writes:

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.

I am confused how he makes this conclusion (bolded). So I know that $n^b$ is made up of a bunch of $7$s and this is its prime factorization. But how can we extend that conclusion to $n^1$?

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    If $n$ had a pruimne factor other than $7$, so would $n^k$. – lulu Mar 29 '21 at 19:19
  • In line with lulu's comment, for illustration, suppose that $$n = (p_1)^{a_1}(p_2)^{a_2}(p_3)^{a_3}\cdots (p_r)^{a_r}.$$ Then $$n^k = (p_1)^{ka_1}(p_2)^{ka_2}(p_3)^{ka_3}\cdots (p_r)^{ka_r}.$$ – user2661923 Mar 29 '21 at 19:24

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The author obtains equation $7^a=n^b$. It follows the prime factorization of $n^b$ is the same as the prime factorization of $n^b$. In order for $n^b$ to be a power of $7$ we must have that $n$ is also a power of $7$. Because if $p$ divides $n$ for a prime other than $7$ then $p$ also divides $n^b$

Asinomás
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$b$ is an integer for starters. Moreover, lets say that $n$ did factor, say with primes $p_1,\dots,p_k$ and exponents $n_1,\dots,n_k$. Then $n^b$ would factor as $p_1^{n_1b},\dots,p_k^{n_kb}$.

But the only prime in the factorization of $n^b$ is 7; so the only prime that can appear in the factorization of $n$ is 7.

Joseph DeGaetani
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