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I am currently doing things related to the Akra–Bazzi theorem. One element in that theorem is the following:

For $n>0$ and sequences of real numbers $a_i, b_i$ of length $n$, where all $a_i>0$ and all $b_i\in (0;1)$, we consider the unique real number $x$ such that $$\sum_{i=1}^n a_i b_i^x = 1$$

My question is: when is $x$ rational/algebraic/transcendental?

For example, for the equation $$\left(\frac{1}{3}\right)^x + \left(\frac{3}{4}\right)^x = 1$$ $x$ is in $(1.1519623;1.1519624)$. It looks like a transcendental number to me, but I have no idea how to show that it is or is not.

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    $\begingroup$ Every number looks like a transcendental number. Also, every number looks like an algebraic irrational, and every number looks like a rational number. $\endgroup$
    – Gerry Myerson
    May 7, 2015 at 12:59
  • $\begingroup$ And certainly every number also looks like a computable number. $\endgroup$
    – Matemáticos Chibchas
    May 7, 2015 at 13:54
  • $\begingroup$ I have no idea what you mean by that. $\endgroup$
    – Manuel Eberl
    May 7, 2015 at 14:40
  • $\begingroup$ Of course the solution of $$ \left(\frac{1}{3}\right)^x + \left(\frac{3}{4}\right)^x = 2 $$ is rational, and the solution of $$ \left(\frac{1}{3}\right)^x + \left(\frac{3}{4}\right)^x = \frac{13}{12} $$ is rational. And of course there are many more like this. So any proof that your number is transcendental seems unlikely to me. See? I can do "seems to me" statements, too. $\endgroup$
    – Gerald Edgar
    May 7, 2015 at 14:54

1 Answer 1

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Schanuel's conjecture implies that if $x$ is an algebraic irrational, $n^x$ for natural numbers $n$ are linearly independent over the rationals: see Will Sawin's answer to this recent question for a proof. It's easy to extend this to positive rationals. Thus Schanuel implies that if the $a_i$ and $b_i$ are rational, $x$ will either be rational or transcendental. So (if you're willing to accept Schanuel) you just have to check for rational solutions, which for something like $(1/3)^x + (3/4)^x = 1$ is not hard.

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  • $\begingroup$ That looks pretty interesting, thanks. However, I'm not entirely sure how to prove that the solution is not rational for a case like that. $\endgroup$
    – Manuel Eberl
    May 8, 2015 at 6:48

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