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Are there integers $a, b > 1$ such that $\pi = \log_a(b)$?

Or equivalently: are there integers $a,b > 1$ such that $a^\pi = b$?

Note that the transcendence of $\pi$ makes this a problem - otherwise the Gelfond-Schneider theorem would tell the answer.

When I asked this question about 10 years ago, I got an answer by Ignat Soroko (Minsk / Belarus) arguing that the problem is likely out of reach for present methods, cf. http://www.gap-system.org/DevelopersPages/StefanKohl/problems/a%5Epi=b.dvi. Is this still the present state?

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Is there a particular reason to consider $\pi$ and not other transcendental numbers? For example, is the problem known for $e$ or $\sum_i 10^{-i!}$? Just for curiosity, can you give an example of a number $x$ such that $a^x=b$ but $x$ is not defined as $\log_a(b)$ (I know this not a precise question)? –  Ricky Jan 10 '13 at 12:10
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Concerning the linked page: as far as I know, $e$ and $\pi$ are not known to be algebraically independent, and in fact, $e+\pi$ and $e\pi$ are still not known to be irrational. What Nesterenko proved is that $\pi$, $\color{red}{e^\pi}$, and $\Gamma(1/4)$ are algebraically independent. –  Emil Jeřábek Jan 10 '13 at 12:43
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@Ricky: Obviously one can ask the question for other transcendental numbers as well - but I think I am maybe not the only one who regards π as a number of particular interest. So far I don't know examples of transcendental numbers x not defined in terms of logarithms such that ax=b for some integers a,b>1. –  Stefan Kohl Jan 10 '13 at 14:36
    
@Emil: Indeed Nesterenko proved 'only' algebraic independence of $\pi$, $e^\pi$ and $\Gamma(\frac{1}{4})$. -- Thank you very much for pointing this out! Also, interesting that transcendence or even only irrationality of $e+\pi$ and $e\pi$ still has not been settled. –  Stefan Kohl Jan 10 '13 at 14:41
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1 Answer

up vote 28 down vote accepted

There is a reason why one can say a little bit more about this question in the case of $\pi$. Because $\pi$ is (essentially) the natural logarithm of a rational number, questions like this are easily derived from Schanuel's conjecture, which states:

If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are complex numbers that are linearly independent over $\mathbb Q$, then the transcendence degree of the field $\mathbb{Q}(\alpha_1, e^{\alpha_1}, \alpha_2, e^{\alpha_2}, \ldots, \alpha_n, e^{\alpha_n})$ over $\mathbb Q$ is at least $n$.

Your proposed equation is $$\pi = \frac{\ln b}{\ln a}\qquad (*)$$

and $(*)$ contradicts the $n=3$ case of Schanuel's conjecture as follows. Take $\alpha_1=\ln a$, $\alpha_2 = \ln b$, and $\alpha_3 = i\pi$. Then $(*)$ implies that $\alpha_1$ and $\alpha_2$ are linearly independent over $\mathbb Q$ (because $\pi$ is irrational), and $\alpha_3$ is trivially linearly independent of $\alpha_1$ and $\alpha_2$ because $\alpha_3$ is purely imaginary and $\alpha_1$ and $\alpha_2$ are real. But $e^{\alpha_1}$, $e^{\alpha_2}$, and $e^{\alpha_3}$ are all rational (even integral) so Schanuel's conjecture implies that $\alpha_1$, $\alpha_2$, and $\alpha_3$ are algebraically independent. This contradicts $(*)$.

The $n=3$ case of Schanuel's conjecture is still open. There are various partial results known (see for example the final chapter of Baker's book Transcendental Number Theory), but I very much doubt that any of these partial results suffice to disprove $(*)$. Note that if we take $n=2$ and $\alpha_1 = \ln \beta_1$ and $\alpha_2 = \ln \beta_2$ for nonzero algebraic numbers $\beta_1$ and $\beta_2$ then we recover the Gelfond–Schneider theorem.

It's a good exercise to derive various statements of this type (e.g., that $e+\pi$ or $\pi^e$ or whatever are transcendental) from Schanuel's conjecture. This way you will be able to figure out on your own whether the statement is likely to be known, or at least be able to approach an expert in transcendental number theory with a more targeted question.

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Dear Timothy, I removed two words from your excellent answer because I feared they could be misinterpreted. If you don't agree with their removal, I apologize, and you should feel free to add them back in. Best wishes, Tom –  Tom Church Jan 11 '13 at 19:05
    
Tom, good catch! –  Timothy Chow Jan 11 '13 at 19:42
    
@Timothy: Good answer! -- I will wait some time whether someone can still tell some more, and if not, I will accept your answer. –  Stefan Kohl Jan 11 '13 at 22:13
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