If $3 y-2 x=4$, determine the value of $\frac{16^{x+1}}{8^{2 y-1}}$.
Using exponent laws, we have

$$
\begin{aligned}
\frac{16^{x+1}}{8^{2 y-1}} & =\frac{\left(2^{4}\right)^{x+1}}{\left(2^{3}\right)^{2 y-1}} \\
& =\frac{2^{4(x+1)}}{2^{3(2 y-1)}} \\
& =2^{(4 x+4)-(6 y-3)} \\
& =2^{4 x-6 y+7}
\end{aligned}
$$

We were given that $3 y-2 x=4$. Multiplying both sides of this equation by -2 leads to $4 x-6 y=-8$. Adding 7 to both sides of this equation gives $4 x-6 y+7=-8+7=-1$.

From above, we have

$$
\frac{16^{x+1}}{8^{2 y-1}}=2^{4 x-6 y+7}=2^{-1}=\frac{1}{2}
$$

ANSWER: $\frac{1}{2}$