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I'm having difficulty with the following problem:

Is there such an $x$ that both $2^{\frac{x}{3}}$ and $3^{\frac{x}{2}}$ are simultaneously rational, for $0<x<1$?

I've tried proving by contradiction but the idea failed. Any help is appreciated!

Yuval Gat
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EngineerInProgress
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  • Sorry I forgot to write one key assumption: $0<x<1$ – EngineerInProgress Feb 21 '19 at 17:22
  • This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. – Carl Mummert Feb 21 '19 at 17:26
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    This is the full question. It's not a part of any other task. It was given as a puzzle on a polish FB math forum. I though I may post it here as I keep trying to solve it ^^ (if this is what you asked for) – EngineerInProgress Feb 21 '19 at 17:29
  • Can you give a link to that fb post? – Bart Michels Feb 21 '19 at 17:31
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    https://www.facebook.com/Delta.czasopismo/photos/a.826894094035013/2325456017512139/?type=3&theater – EngineerInProgress Feb 21 '19 at 17:34
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    Maybe easier: Is there a $y$ such that $4^y$ and $27^y$ are both rational, with $0 < y < \frac16$. – Paul Feb 21 '19 at 23:03
  • Given the comments to MSE question 1375825 "Are logarithms of prime numbers algebraically independent?" almost nothing is known about such questions. – Somos Feb 25 '19 at 12:06
  • @Somos, I don't see why that's relevant. The case of linear independence for natural logarithms of primes is known, and that seems much closer to this question than the algebraic independence mentioned in that question. Also note the question has an easy affirmative answer if you remove the restriction that $x<1$, so your objection would have to somehow only apply to that. – Cheerful Parsnip Feb 26 '19 at 05:15
  • The problem can easily be reduced to the Diophantine equation $\left(\frac{r}{s}\right)^{\log{4}}= \left(\frac{p}{q}\right)^{\log{27}}$ for the positive integers $p,q,r,s$ subject to the conditions (1) $1\lt \frac{r}{s}\lt \sqrt{3}$ and (2) $1\lt \frac{p}{q}\lt2^{\frac{1}{3}}$ – Dr. Wolfgang Hintze Feb 26 '19 at 16:00
  • Let me make a bold statement: I doubt that there is a solution because I don't see that the $\frac{\log{2}}{\log{3}}$-th power of a rational number can be a rational number. – Dr. Wolfgang Hintze Feb 26 '19 at 16:14
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    @Dr.WolfgangHintze If I’m not mistaken, $3^\frac {\log 2}{\log 3}=2$. – Alex Ravsky Feb 27 '19 at 21:08
  • @ Alex Ravsky Oops. Thank you for pointing out the error in my statement which was far too bold ;-) – Dr. Wolfgang Hintze Feb 28 '19 at 09:52
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    Error with earlier comment on GS : Note that it shows that $x$ cannot be algebraic and irrational. It can be transcendental, however. Also, I think no rational $x$ works, since $2^{\frac x3}$ for $0<x < 1$ rational ensures that $\frac x3$ will always have a non-trivial denominator, so we get a surd which is irrational.Therefore, any $x$ for which $2^{\frac x3}$ is rational is transcendental, but then that's a not so useful statement in light of how large and uncharacterizable the transcendental numbers are. – Sarvesh Ravichandran Iyer Feb 28 '19 at 17:11
  • $\exists y \notin \mathbb Z . 4^y, 27^y \in \mathbb Q \Leftrightarrow \exists y \notin \mathbb Z . y \log 4, y \log 27 \in \mathrm{span}{\mathbb Z}{\log p : p\text{ prime}} \Leftrightarrow \exists z \notin (\log 4\log 27)\cdot\mathbb Z . z \in \mathrm{span}{\mathbb Z}{\log 4\log p : p\text{ prime}} \cap \mathrm{span}_{\mathbb Z}{\log 27\log p : p\text{ prime}} \Leftrightarrow$ ${\log 4\log p : p\text{ prime}} \cup {\log 27\log p : p\text{ prime}}$ is dependent over $\mathbb Z$. So the relevant question seems to be whether ${\log p :p\text{ prime}}$ is "quadratically independent". – eccheng Mar 02 '19 at 12:41

1 Answers1

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Negative answer follows from the four exponentials conjecture. Indeed, it's well-known that $\frac{\log 2}{3},\frac{\log 3}{2}$ are linearly independent over $\mathbb Q$ (a relation between them would imply an equality of the form $2^a=3^b$ for $a,b\in\mathbb N_+$ which clearly cannot happen). Further, it's easy to see $x$ must be irrational too, so that $1,x$ are linearly independent over $\mathbb Q$. Now the conjecture mentioned implies that one of the following numbers is transcendental: $$\begin{align*} e^{1\cdot\frac{\log 2}{3}} &=2^{1/3},\\ e^{1\cdot\frac{\log 3}{2}} &=3^{1/2},\\ e^{x\cdot\frac{\log 2}{3}} &=2^{x/3},\\ e^{x\cdot\frac{\log 3}{2}} &=3^{x/2}.\end{align*}$$ The first two numbers are clearly algebraic, so we conclude we can't have the latter two simultaneously rational.

It should be noted that the four exponentials conjecture is still wide open, despite the fact there are some very similar results which have been proven. A related question, asking when $2^x,3^x$ can be both integral, is also well-known and open.

Wojowu
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