27. Find the number of positive integers $x$, where $x \neq 9$, such that
$$
\log _{\frac{x}{9}}\left(\frac{x^{2}}{3}\right)<6+\log _{3}\left(\frac{9}{x}\right) .
$$
27. Answer: 223
$$
\begin{aligned}
& \log _{\frac{x}{9}}\left(\frac{x^{2}}{3}\right)<6+\log _{3}\left(\frac{9}{x}\right) \\
\Leftrightarrow & \frac{\log _{3}\left(\frac{x^{2}}{3}\right)}{\log _{3}\left(\frac{x}{9}\right)}<6+\log _{3} 9-\log _{3} x \\
\Leftrightarrow & \frac{\log _{3} x^{2}-\log _{3} 3}{\log _{3} x-\log _{3} 9}<6+\log _{3} 9-\log _{3} x .
\end{aligned}
$$

Let $u=\log _{3} x$. Then the inequality becomes $\frac{2 u-1}{u-2}<8-u$, which is equivalent to $\frac{u^{2}-8 u+15}{u-2}<0$. Solving the inequality gives $u<2$ or $3<u<5$, that is, $\log _{3} x<2$ or $3<\log _{3} x<5$. It follows that $0<x<9$ or $27<x<243$. Hence there are 223 such integers.