9・15 If the relation is
$$
\log _{3}\left[\log _{\frac{1}{2}}\left(\log _{2} x\right)\right]=\log _{3}\left[\log _{\frac{1}{3}}\left(\log _{3} y\right)\right]=\log _{3}\left[\log _{\frac{1}{5}}\left(\log _{5} z\right)\right]=0 .
$$

then the size relationship of $x, y, z$ is
(A) $x<y<z$.
(B) $y<z<x$.
(C) $z<x<y$.
(D) $z<y<x$.
(China High School Mathematics League, 1982; China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1988)
[Solution 1] From $\log _{3}\left[\log _{\frac{1}{2}}\left(\log _{2} x\right)\right]=0$, we get
$$
\log _{\frac{1}{2}}\left(\log _{2} x\right)=1, \log _{2} x=\frac{1}{2},
$$

so
$$
x=2^{\frac{1}{2}}.
$$

Similarly, we get $y=3^{\frac{1}{3}}, \quad z=5^{\frac{1}{5}}$. Converting to the same root, we have
$$
\begin{array}{l} 
x=\sqrt{2}=\sqrt[30]{2^{15}}=\sqrt[30]{32768}; \\
y=\sqrt[3]{3}=\sqrt[30]{3^{10}}=\sqrt[30]{59049}; \\
z=\sqrt[5]{5}=\sqrt[30]{5^{6}}=\sqrt[30]{15625}. \\
\therefore \quad z<x<y.
\end{array}
$$

Therefore, the answer is $(C)$.
[Solution 2] By the properties of logarithms and the given conditions, we have
$$
x=2^{\frac{1}{2}}, \quad y=3^{\frac{1}{3}}, \quad z=5^{\frac{1}{5}}.
$$

Also, from $\left\{\begin{array}{l}x^{6}=2^{3}=8, \\ y^{6}=3^{2}=9,\end{array}\right.$ we get $x<y$. And from $\left\{\begin{array}{l}x^{10}=2^{5}=32, \\ z^{10}=5^{2}=25,\end{array}\right.$ we get $z<x$.
In summary, we have
$$
z<x<y.
$$

Therefore, the answer is $(C)$.