Does there exist a real number $r$ such that $\pi^r$ is a rational number? Clearly, $(\sqrt{2})^2=2$ etc. Is there any real number such that my question is true?
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$\begingroup$ @Gibbs, editing and removing the word "real" from real number is a major decision... $\endgroup$– zwimCommented Dec 10, 2017 at 17:39
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$\begingroup$ I did not touch that part, I just fixed the math syntax. $\endgroup$– GibbsCommented Dec 10, 2017 at 17:41
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$\begingroup$ You are right, it seems I 'corrected' it, but I actually didn't... I always fix only the math syntax... I am sorry, I edit it again. Thanks for the comment. $\endgroup$– GibbsCommented Dec 10, 2017 at 17:43
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$\begingroup$ To state the obvious, $r=0$ works. Presumably you meant to exclude that but details matter. Did you have any other conditions in mind for $r$? $\endgroup$– luluCommented Dec 10, 2017 at 17:43
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1 Answer
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What you are looking for is a logarithm with base $\pi$.
Take $$\pi^r=n,\;\;\;\; \log_{\pi}(n)=r$$
Using change of base: $$r=\frac{\ln(n)}{\ln\pi}$$
Since the logarithm is defined on the positive real line, there exists some number $r\in\Bbb R$ which satisfies the condition.
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$\begingroup$ It dawned on me a few min after I wrote this that I was really hoping there might be a rational number r. What I was asking is just like we can define any number as e to some power. Of course, all these values of $e^x$ for any value of x were calculated years ago like sinx, cosx etc. But, these are really approximations except for a few values like sinx = 1/2, 1 etc, etc.. – discountbrainsurgery yesterday $\endgroup$ Commented Dec 12, 2017 at 5:01