China Mathematical Competition

Note that $f(x) = |\log_3(\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\infty)$.
By the conditions satisfied by $a, b, c$, we know that $0 < a < b < 9 < c$ and
$$\log_3\left(\frac{9}{a}\right) = 2\log_3\left(\frac{9}{b}\right) = 2\log_3\left(\frac{c}{9}\right).$$
Therefore,
$$\begin{aligned} \log_3\left(\frac{ac}{b}\right) &= \log_3\left(9 \cdot \frac{a}{9} \cdot \frac{9}{b} \cdot \frac{c}{9}\right) \\ &= 2 - \log_3\left(\frac{9}{a}\right) + \log_3\left(\frac{9}{b}\right) + \log_3\left(\frac{c}{9}\right) \\ &= 2, \end{aligned}$$
namely, $\frac{ac}{b} = 3^2 = 9$.