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I am looking at the rationality of $\log_ba$ when at least $a$ or $b$ is irrational such that $a,b>1$ and neither of $a$ or $b$ is of the form $\left(\frac{e}{f}\right)^{(g/d)}$, where $d,e,f,\text{ and }g\in\mathbb{N}$.

I know how to check irrationality when both of $a$ and $b$ are rational, one of the beautiful answer by Gone is here.

I am done with the case when exactly one of them is irrational.
Here my version of proof:

W.L.O.G, Assume $a$ is irrational.
Suppose that $$\log_ba=p/q\enspace\text{ where } p,q\in\mathbb{N}\\\implies a^q=b^p$$ but $a^q$ is irrational and $b^p$ is rational. Hence, contradiction.

Also, if $a=b^c$, then it depends on $c$.

Finally, when $a$ and $b$ are irrational and $a\neq b^c$.

I have no idea about it. Can anyone help me with this? Also, is my above proof correct? If incorrect, please do provide the correct one. Thanks.

Kumar
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  • If $\log_b a = c$, then $a = b^c$. So what do mean by "$a \ne b^c$"? – Robert Israel May 08 '20 at 15:36
  • @RobertIsrael I want to say the other way round. It's if $a=b^c$ then $\log_ba=c$. – Kumar May 08 '20 at 15:40
  • In your proof where exactly one is irrational, how do you know $a$ is irrational $\implies$ $a^q$ is irrational? Try $b=2, a= \sqrt{2}$. – healynr May 08 '20 at 15:44
  • Why is $a^p$ irrational? Say, you put $a=\sqrt{2}$ and $q=\text{even}$? – AryanSonwatikar May 08 '20 at 15:45
  • @Kumar yes, sorry I missed that. – AryanSonwatikar May 08 '20 at 15:59
  • @Kumar When you say $a \not = b^c$, is $c$ an integer or a rational number? – healynr May 08 '20 at 16:15
  • @healynr While, Thinking about it, I had in my mind that $c\in\mathbb{R}$ with simple examples of rational numbers and some irrational such as $\pi$, but as you ask the question, I am unable to decide it as I have no idea whether we can express $\mathbb{R}$ as a real power of some fixed irrational number? So, for cautionary purposes, I suppose we need to work with rational numbers only. :( – Kumar May 08 '20 at 16:41
  • @Kumar Well if $a \not = b^c$ for some rational number $c$, then of course $\log_b a \not \in \mathbb{Q}$ since if it were, then $a = b^{\log_b a}$! – healynr May 08 '20 at 16:44

1 Answers1

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If $a \not = b^c$ for $c \in \mathbb{Q}$, then $\log_b a = c \not \in \mathbb{Q}$. Now if we reduce the assumption to $c \in \mathbb{Z}$:

Let $a=\sqrt{\pi}$ and $b=\pi$. Then since neither $a,b$ are algebraic nor rational, they fit your assumption that neither equal $\left(\frac{e}{f}\right)^{(g/d)}$. $$\log_{b}a=\log_{\pi}\sqrt{\pi}=\frac{1}{2} \in \mathbb{Q}$$

healynr
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