9.163. $\frac{\lg 7-\lg \left(-8 x-x^{2}\right)}{\lg (x+3)}>0$.

9.163. $\frac{\log 7-\log \left(-8 x-x^{2}\right)}{\log (x+3)}>0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}-8 x-x^{2}>0, \\ x+3>0, \\ \lg (x+3) \neq 0\end{array} \Leftrightarrow\left[\begin{array}{l}-30, \log _{x+3} \frac{7}{-8 x-x^{2}}>0$. The obtained

inequality is equivalent to the following two systems of inequalities:

1) $\left\{\begin{array}{l}00\end{array} \Leftrightarrow\left\{\begin{array}{l}-30\end{array} \Leftrightarrow\right.\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}-30 .\end{array}\right.$

From here, using the interval method, we find $-31, \\ \frac{7}{-8 x-x^{2}}>1\end{array} \Leftrightarrow\left\{\begin{array}{l}x>-2, \\ 1+\frac{7}{x^{2}+8 x}-2, \\ \frac{x^{2}+8 x+7}{x^{2}+8 x}-2, \\ (x+8) x(x+7)(x+1)<0 .\end{array}\right.$

Using the number line, we find $-1<x<0$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0679.jpg?height=113&width=501&top_left_y=1564&top_left_x=468)

Combining both cases, we get $\left[\begin{array}{l}-3<x<-2, \\ -1<x<0 .\end{array}\right.$

Answer: $x \in(-3 ;-2) \cup(-1 ; 0)$.