A graphical inequality and exponential substitution

Sketch $y=|x/2+3|-(3-x^2)$. The graph meets the horizontal axis at $-1/2$ and $0$, lies below it between these points and above it outside. Thus $x<-1/2$ or $x>0$.

Sketch $y=|x/2+3|$ and $y=3-x^2$. The absolute-value graph has vertex $(-6,0)$ and passes through $(0,3)$. The parabola has vertex $(0,3)$ and horizontal intercepts $\pm\sqrt3$. Their intersections have inputs $-1/2$ and $0$. The absolute-value graph lies above the parabola for $x<-1/2$ or $x>0$.

Replace the original input with $-e^x$. Since $|-e^x/2+3|=|e^x/2-3|$ and $(-e^x)^2=e^{2x}$, obtain $-e^x>0$ or $-e^x<-1/2$. Reject the first because $-e^x<0$ for every real $x$. The second gives $e^x>1/2$, hence $x>\ln(1/2)$.