7.045. $\log _{5}(3 x-11)+\log _{5}(x-27)=3+\log _{5} 8$.
Solution.

Domain of definition: $\left\{\begin{array}{l}3 x-11>0, \\ x-27>0,\end{array} \quad x>27\right.$.

## We have

$$
\begin{aligned}
& \log _{5}(3 x-11)+\log _{5}(x-27)=\log _{5} 125+\log _{5} 8 \\
& \log _{5}(3 x-11) \cdot(x-27)=\log _{5}(125 \cdot 8) \quad(3 x-11)(x-27)=125 \cdot 8 \\
& 3 x^{2}-92 x-703=0
\end{aligned}
$$

from which we find $x_{1}=37, x_{2}=-\frac{19}{3} ; x_{2}=-\frac{19}{3}$ does not satisfy the domain of definition.

Answer: 37.