7.187. $x^{2} \cdot \log _{x} 27 \cdot \log _{9} x=x+4$.
## Solution.

Domain of definition: $0<x \neq 1$.

Switch to base 3, then

$$
\frac{3 x^{2}}{\log _{3} x} \cdot \frac{\log _{3} x}{2}=x+4 \Leftrightarrow 3 x^{2}-2 x-8=0
$$

from which $x_{1}=2, x_{2}=-\frac{4}{3} ; x_{2}=-\frac{4}{3}$ does not satisfy the domain of definition.
Answer: 2.