Does $f=o(e^{x})$ yield $ \exists \rho (0<\rho<1): f=o(e^{\rho x})$?

Question

Does $f=o(e^{x})$ yield $ \exists \rho(0<\rho<1):$ $f=o(e^{\rho x})$?

Intuitively, I think this is true since if there is no such $\rho$, it seems that $f=\Theta(e^ {x})$ which is a contradiction. I don't know how to show this.