10. (This question is worth 20 points) Given the function $f(x)$ defined on $\mathbf{R}^{+}$ as
$$
f(x)=\left\{\begin{array}{ll}
\left|\log _{3} x-1\right|, & 09 .
\end{array}\right.
$$

Let $a, b, c$ be three distinct real numbers, satisfying $f(a)=f(b)=f(c)$, find the range of values for $a b c$.
Solution: Without loss of generality, let $a<b<c$. Since $f(x)$ is strictly decreasing on $(0,3]$, strictly increasing on $[3,9]$, and strictly decreasing on $[9,+\infty)$, and given $f(3)=0, f(9)=1$, it follows from the graph that
$$
a \in(0,3), b \in(3,9), c \in(9,+\infty),
$$

and $f(a)=f(b)=f(c) \in(0,1)$. $\qquad$
From $f(a)=f(b)$, we get
$$
1-\log _{3} a=\log _{3} b-1,
$$

which means $\log _{3} a+\log _{3} b=2$, so $a b=3^{2}=9$. Thus, $a b c=9 c$. $\qquad$ Also,
$$
0<f(c)=4-\sqrt{c}<1,
$$
$\qquad$
so $c \in(9,16)$. Therefore, $a b c=9 c \in(81,144)$.
Hence, the range of $a b c$ is $(81,144)$. $\qquad$
Note: For any $r \in(81,144)$, let $c_{0}=\frac{r}{9}$, then $c_{0} \in(9,16)$, so $f\left(c_{0}\right) \in(0,1)$. Draw a line $l$ parallel to the $x$-axis through the point $\left(c_{0}, f\left(c_{0}\right)\right)$, then $l$ intersects the graph of $f(x)$ at two other points $(a, f(a))$, $(b, f(b))$ (where $a \in(0,3), b \in(3,9)$), satisfying $f(a)=f(b)=f(c)$, and $a b=9$, thus $a b c=r$.