7.114. $\lg (x(x+9))+\lg \frac{x+9}{x}=0$.
## Solution.

Domain of definition: $x(x+9)>0, x \in(-\infty ;-9) \cup(0 ; \infty)$.

We have $\lg \frac{x(x+9)(x+9)}{x}=0$, from which $(x+9)^{2}=1$. Then $(x+9)_{1}=-1$, $x_{1}=-10$ or $(x+9)_{2}=1, x_{2}=-8 ; x_{2}=-8$ does not fit the domain of definition.

Answer: -10.