Answer
$(2, 4) \cup (7, 9)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a polynomial function.
$\log(x-2)+\log(9-x)<1,$
$\log{(x-2)(9-x)} < 1,$
$\log{(-x^2+11x-18)} < 1,$
$-x^2+11x-18<10,$
$-x^2+11x-28<0,$
$-x^2+4x+7x-28<0,$
$-x(x-4)+7(x-4)<0,$
$(-x+7)(x-4)<0,$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(-x+7)(x-4)<0,$
$x=7$ or $x=4$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll} Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\ & &(-a+7)(a-4)<0& \\ (-\infty, 4) & 0 & (+)(-)=(-) & T\\ (4, 7) & 5 & (+)(+)=(+) & F\\ (7,\infty) & 10 & (-)(+)=(-) & T \end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
-Domain of logarithmic function is $(0,\infty)$.
Therefore, $x-2\gt0, x\gt2$ and $9-x\gt0, x\lt9$.
Solution set: $(2, 4) \cup (7, 9)$
