May $y=e^x$ be satisfied with both $x$ and $y$ $\in$ $\mathbb{Z}^+$?

Question

May $y=e^x$ be satisfied with both $x$ and $y$ be positive integers?

I think it is not possible as $e$ ,a transcendental number, when multiplied by itself would never result in rational number.

Am I right?

Answer 1

Yes, otherwise there would be a polynomial $X^a-y$ for which $e$ would be a root.

Answer 2

Suppose yes! Then $$e=\sqrt[x]{\phantom{(}y\phantom{)}}$$ where both $x,y$ are positive integers. Then $e$ is a solution to the polynomial equation $$X^x-y=0.$$ As $e$ is transcendental, this is a contradiction. Hence there a no such integers.

Answer 3

Your equation cannot be satisfied according to the Lindemann–Weierstrass theorem.

However, it is not generally true that a product of two transcendental numbers must be transcendental. If that were the case, it wouldn't be an open question whether $\pi \mathrm e$ were transcendental or not.