Is there a proof that logarithm whose anti-logarithm is not a natural power of the base is not a rational number?

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To write it simplier (at least for me):

Is there a proof that

$\forall a,b\in\mathbb{N}_+\backslash{\{1\}},b>a, \text{a is prime}, \forall k\in\mathbb{N}, b \neq a^k: \\ \log_a b\notin \mathbb{Q}$

so for example (that this question arose from)

Is $\log_2 b$ a rational number only when b ($b\in\mathbb{N}_+$) is a natural power of 2?

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We'll prove the final version by contrapositive. Suppose that $\log_a(b)\in\mathbb{Q}$. Then there exists $q,p\in\mathbb{Z}$ such that $(q,p)=1$ and

$$a^{p/q} = b.$$

Since $b > a > 1$ we can require that $p > q > 0$. Now raise both sides to the $q$ to obtain

$$a^p = b^q.$$

Since $a$ is prime the left hand side is the prime factorization of $a^p$ and so the prime factorization of the right hand side must also only contain powers of $a$ which means $b=a^k$ for some $k\in\mathbb{N}$.