13. [9] Suppose $x, y$, and $z$ are real numbers greater than 1 such that
$$
\begin{aligned}
x^{\log _{y} z} & =2, \\
y^{\log _{y} x} & =4, \text { and } \\
z^{\log _{x} y} & =8 .
\end{aligned}
$$

Compute $\log _{x} y$.
Answer: $\sqrt{3}$
Solution: Taking $\log _{2}$ both sides of the first equation gives
$$
\begin{array}{c}
\log _{2} x \log _{y} z=1 \\
\frac{\log _{2} x \log _{2} z}{\log _{2} y}=1 .
\end{array}
$$

Performing similar manipulations on other two equations, we get
$$
\begin{array}{l}
\frac{\log _{2} x \log _{2} z}{\log _{2} y}=1 \\
\frac{\log _{2} y \log _{2} x}{\log _{2} z}=2 \\
\frac{\log _{2} z \log _{2} y}{\log _{2} x}=3 .
\end{array}
$$

Multiplying the first and second equation gives $\left(\log _{2} x\right)^{2}=2$ or $\log _{2} x= \pm \sqrt{2}$. Multiplying the second and third equation gives $\left(\log _{2} y\right)^{2}=6$ or $\log _{2} y= \pm \sqrt{6}$. Thus, we have
$$
\log _{x} y=\frac{\log _{2} y}{\log _{2} x}= \pm \frac{\sqrt{6}}{\sqrt{2}}= \pm \sqrt{3}
$$