Is there such an $x$ that both $2^{\frac{x}{3}}$ and $3^{\frac{x}{2}}$ are simultaneously rational? I'm having difficulty with the following problem:

Is there such an $x$ that both $2^{\frac{x}{3}}$ and $3^{\frac{x}{2}}$ are simultaneously rational, for $0<x<1$?

I've tried proving by contradiction but the idea failed. Any help is appreciated!
 A: Negative answer follows from the four exponentials conjecture. Indeed, it's well-known that $\frac{\log 2}{3},\frac{\log 3}{2}$ are linearly independent over $\mathbb Q$ (a relation between them would imply an equality of the form $2^a=3^b$ for $a,b\in\mathbb N_+$ which clearly cannot happen). Further, it's easy to see $x$ must be irrational too, so that $1,x$ are linearly independent over $\mathbb Q$. Now the conjecture mentioned implies that one of the following numbers is transcendental:
$$\begin{align*}
e^{1\cdot\frac{\log 2}{3}} &=2^{1/3},\\
e^{1\cdot\frac{\log 3}{2}} &=3^{1/2},\\
e^{x\cdot\frac{\log 2}{3}} &=2^{x/3},\\
e^{x\cdot\frac{\log 3}{2}} &=3^{x/2}.\end{align*}$$
The first two numbers are clearly algebraic, so we conclude we can't have the latter two simultaneously rational.
It should be noted that the four exponentials conjecture is still wide open, despite the fact there are some very similar results which have been proven. A related question, asking when $2^x,3^x$ can be both integral, is also well-known and open.
