2. Given $x, y, z > 1$, satisfying
$$
\log _{x} 2+\log _{y} 4+\log _{z} 8=1 \text {. }
$$

Then the minimum value of $x y^{2} z^{3}$ is $\qquad$
$2.2^{36}$.
Let $a=\log _{x} 2, b=\log _{y} 4, c=\log _{z} 8$. Then $a+b+c=1$.
Therefore, $x y^{2} z^{3}=2^{\frac{1}{a}} \times 4^{\frac{2}{b}} \times 8^{\frac{3}{c}}=2^{\frac{1}{a}+\frac{4}{b}+\frac{9}{c}}$ $=2^{\left(\frac{1}{a}+\frac{4}{b}+\frac{9}{c}\right)(a+b+c)} \geqslant 2^{(1+2+3)^{2}}=2^{36}$.