5. Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$, to make $\log _{\frac{x_{0}}{x_{1}}} 1993+$ $\log \frac{x_{1}}{x_{2}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993>k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is
5. Rewrite the original inequality as
$$
\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \geqslant k \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}}
$$

i.e.,
$$
\begin{array}{c}
\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \geqslant k \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}} \\
\text { Let } t_{1}=\log _{1993} \frac{x_{0}}{x_{1}}, \quad t_{2}=\log _{1993} \frac{x_{1}}{x_{2}}, \quad t_{3}=\log _{1993} \frac{x_{2}}{x_{3}},
\end{array}
$$

Thus, the original inequality becomes
$$
\begin{array}{l}
\left(t_{1}+t_{2}+t_{3}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}\right) \geqslant k . \\
\left(t_{1}+t_{2}+t_{3}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}\right) \\
\geqslant 3 \sqrt[3]{t_{1} \cdot t_{2} \cdot t_{3}} \cdot 3 \sqrt[3]{\frac{1}{t_{1}} \cdot \frac{1}{t_{2}} \cdot \frac{1}{t_{3}}}=9,
\end{array}
$$

However,

the equality holds if and only if $t_{1}=t_{2}=t_{3}$ (i.e., $x_{0}, x_{1}, x_{2}, x_{3}$ form a geometric sequence), $\therefore \quad k_{\max }=9$.