Is it true that $\log_2 13$ is irrational?

Let $x=\log_2 13\implies 2^x=13$.

So, it will be an irrational number, if not,$$x=\frac p q$$

and $$2^{\frac p q}=13$$

$$\implies 2^p=13^{q}$$

Since, $13$ is a prime number, $2^p$ divides $13^q$.

So, $2$ divides $13$, which is absurd.

Is this reason worthy? Can you give some other proofs for this?

  • 13
    $\begingroup$ That's a nice and elegant proof. You can extend it to $$\log_a b \notin \mathbb Q \qquad \forall a,b\in\mathbb N\text{ s.t. } \gcd(a,b) = 1$$ $\endgroup$
    – AlexR
    Jan 16 '15 at 14:22
  • $\begingroup$ Thank you.. Is there any other proofs??? $\endgroup$
    – David
    Jan 16 '15 at 14:25
  • 2
    $\begingroup$ I don't immediately see how else to tackle the problem, since $2^x = 13$ is an exponential function, so solving it would resort to $\log$ of some basis, but since we want to prove a property of $\log$, that's not going to help us much. Maybe you can reduce it to some other $\log_a b \notin\mathbb Q$ statement this way. $\endgroup$
    – AlexR
    Jan 16 '15 at 14:29
  • $\begingroup$ You could use Gelfond-Schneider theorem, but it's not really necessary to resort to such heavy artillery. $\endgroup$ Apr 13 '15 at 8:19

You are done in your solution at the step where you concluded that $2^p = 13^q$. You only need to quote Fundamental Theorem of Arithmetic after that step.


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