jmc

We write $9$ as a power of $3$, and we have $9^{x+3}=(3^2{)}^{x+3}=3^{2(x+3)}=3^{2x+6}$. Therefore, the original equation is $3^{x+8}=3^{2x+6}$. Equating the two exponents, we get $x+8=2x+6$. Solving, we get that $x=\boxed{2}$.