Exercise 1.3.29 (If $\log (m)/(\log (n)$ is rational, $m = c^a, n = c^b$ for some integer $c$)

Question

Suppose that $m$ and $n$ are integers $>1$, and that $(\log m)/(\log n)$ is rational, say equal to $a/b$ with $(a,b) = 1$. Show that there must be an integer $c$ such that $m = c^a, n = c^b$.

Richard Ganaye · Accepted Answer

\begin{proof} By hypothesis,
$$\frac{\log(m)}{\log(n)} = \frac{a}{b},\qquad a \wedge b = 1.$$
Therefore $$m^b = n^a.$$
Since $a \wedge b = 1$, there are integers $\lambda, \mu$ such that $\lambda a + \mu b = 1$. Then
\begin{align*}
 m &= m^{\lambda a + \mu b}\
 &= m^{\lambda a} (m^b) ^\mu\
 &=m^{\lambda a} (n^a)^\mu\
 &= (m^\lambda n^\mu)^a\
 &= c^a,
\end{align*}
where $c = m^\lambda n ^\mu$.

Similarly,
\begin{align*}
n&= n^{\lambda a + \mu b}\
&= (n^a)^\lambda n ^{\mu b}\
&= (m^{b})^\lambda n ^{\mu b}\
&= (m^\lambda n ^\mu)^b\
&=c^b. 
\end{align*}
There exists an integer $c$ such that $m = c^a, n = c^b$.

\end{proof}

Exercise 1.3.29 (If $\log (m)/(\log (n)$ is rational, $m = c^a, n = c^b$ for some integer $c$)

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Exercise 1.3.29 (If $\log (m)/(\log (n)$ is rational, $m = c^a, n = c^b$ for some integer $c$)

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