jmc

First, we recognize that $2^{x+1}+6=2(2^x+3)$: $$\frac{2(66-2^x)}{2(2^x+3)}=\frac{4-2^x}{2(2^x+3)}$$Then, we expand and collect like terms: $$\frac{128-2^x}{2(2^x+3)}=0$$This equation can only be true when $2^x=128$, which indicates that $x=\boxed{7}$.