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I had this doubt in the question. First, I took log both sides to get $a \log x=b \log y$. Since a does not divide b, a must divide log y. So log y can be written as some $\log y=am$, when we get $y= (10^m)^a$ which is of the form $y=n^a$. Is this correct or there is some flaw with the argument?

Bill Dubuque
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Principia Mathematica
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    We cannot talk about divisibility if $\log y$ is not an integer. – player3236 Feb 17 '21 at 05:17
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    Is it given that $x,y$ are themselves integers? – Mike Feb 17 '21 at 05:18
  • Yes they are integers – Principia Mathematica Feb 17 '21 at 05:27
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    You should avoid using $\log$ in problems involving integers. The key here is that if you have numbers $a,b$ such that $\gcd(a,b) = 1$, then there exist $p,q \in \mathbb{Z}$ such that $ap + bq = 1$. – Naweed G. Seldon Feb 17 '21 at 06:06
  • Hint $\ $ It's a multiplicative form of the following basic theorem about fractions $$aj!+!bk=1,\ {xa=yb},\Rightarrow, \bbox[5px,border:1px solid red]{\dfrac{y}x = \dfrac{a}b:\Rightarrow\begin{align},y = na\ x = nb\end{align}}\ \ \ {\rm for\ some},\ n\in\Bbb Z\qquad $$

    We can mechanically translate a proof of that into the proof you seek (see one of my answers in the linked dupe for details).

     – Bill Dubuque Feb 17 '21 at 09:14

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