Put $ x=\frac{p}{q}$ with $ p,q$ relatively prime, and $ \lfloor x\rfloor=n$. Then we want that $ 2p^{n}=9q^{n}$, so $ 2|q$, for $ n>1$ this means the RHS has more than one factor $ 2$ so $ 2|p$, contradiction with relative prime. So $ n=1$ (since $ n=0$ is impossible), so $ 2p=9q$, so $ x=\frac{9}2$, but then $ n>1$, contradiction.

Another way (though not as good): Suppose $x$ is a positive rational. If $x<1 $ then $x^{\lfloor x\rfloor } = x^0 = 1 \neq 9/2 $. If $1\leq x < 2 $, $x^{\lfloor x\rfloor } = x^1= x \neq 9/2 $ since $9/2 > 2$. If $ 2\leq x < 3 $ then $x^{\lfloor x\rfloor } = x^2 \neq 9/2 $ since $3/\sqrt{2}$ is irrational. If $ 3\leq x $ then $ x^{\lfloor x\rfloor } \geq 3^3 = 27 > 9/2 $. So there are no positive rational solutions to $x^{\lfloor x\rfloor }= \frac{9}{2}$.

$ My $ $ solution $ $ (a,b)=1 $ $ x=\frac{a}{b} $ $ \lfloor x\rfloor=c $ $$ x^{\lfloor x\rfloor}=\frac{a^c}{b^c}=\frac{9}{2}\implies a^c=9,b^c=2\implies c=1,a=9,b=2$$$ Then $ $ \lfloor x\rfloor=1,x=\frac{9}{2} $ No solution