Eliminate exponential and trigonometric parameters

$y=e^{3t}=(e^t)^3=x^3$, with $x>0$.

$t=e^x$, so $y=\sqrt{e^x}=e^{x/2}$. Every real $x$ is possible; there is no further restriction on $x$.

$\sec\theta=(x-5)/4$ and $\tan\theta=y-1$. Since $\sec^2\theta-\tan^2\theta=1$, $(x-5)^2/16-(y-1)^2=1$. The secant requires $\cos\theta\ne0$ and has values $\sec\theta\le-1$ or $\sec\theta\ge1$. Hence $x\le1$ or $x\ge9$.