13. A sequence is given by $x_{1}=2, x_{n+1}=\frac{x_{n} x_{n}}{32^{n-1}}$. What is $\frac{x_{4}}{128}$ ?
SolUTION
128

We determine each term in sequence.
$$
x_{2}=\frac{2^{2}}{32^{0}}=\frac{4}{1}=4 ; x_{3}=\frac{4^{4}}{32^{1}}=\frac{2^{8}}{2^{5}}=2^{3}=8 ; x_{4}=\frac{8^{8}}{32^{2}}=\frac{2^{24}}{2^{10}}=2^{14} \text {. }
$$

Therefore, $\frac{x_{4}}{128}=2^{14} \div 2^{7}=2^{7}=128$.