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If $\log_{b}N$ is rational, is there a set of values to which $b$ and $N$ must belong? Is there a set of values to which $b$ and $N$ cannot belong?

Further, if it is presupposed that $b$ and $N$ are integers, how does that change the answers to the previous two questions?

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3 Answers 3

up vote 1 down vote accepted

If $b$ and $N$ are positive integers, the necessary and sufficient condition is that they are both integer powers of the same positive integer. That is, $b = c^m$ and $N = c^n$ where $\log_b N = n/m$ in lowest terms.

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That's what I thought, but I wasn't sure if it was necessary. I don't suppose you could humor me with a proof of the necessity? –  tel Apr 17 '12 at 23:56
Consider the prime factorization of $b$. If $p^d$ is the largest power of prime $p$ that divides $b$, then $p^{nd}$ is the largest power of $p$ that divides $b^n = N^m$. But then $d$ must be divisible by $m$. Conclude that $b$ is the $m$'th power of an integer. –  Robert Israel Apr 18 '12 at 0:13

Partial answer to the first part:

Assuming $b,N>0$, if $\log_b N=q$ is rational then $b^q=N$ and equally $b=\sqrt[q]N$ so $\log_N b=q^{-1}$ which is also rational. Therefore: $$\log_b N\in\mathbb Q\iff b\in\{N^q\mid q\in\mathbb Q\}\iff N\in\{b^q\mid q\in\mathbb Q\}.$$

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If $\log_{b}^{N}$ is rational, so.

$\log_{b}^{N}=x$ with $x \in \mathbb{Q}$, so the answer tuples should be:

$(b,N)=\forall x \in \mathbb{Q}\quad(b,b^x)$.

But now is needed remove the bound of log. $b\ne1,b>0$ and $N>0$.

I think this could help you to finish the first part.

The second part, about, integers. We can see, we can write the answers using b, so, we know $b$ belongs integers and $N$ is the form $b^x$.

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