$7.24 \log _{3}\left(3^{x^{2}-13 x+28}+\frac{2}{9}\right)=\log _{5} 0.2$.
7.24 Note that the expression under the logarithm sign in the left part is positive for all $x$. Let's simplify the right part:

$$
\log _{5} 0.2=\log _{5} \frac{1}{5}=\log _{5} 1-\log _{5} 5=-1=\log _{3} 3^{-1}
$$

Thus,

$$
\log _{3}\left(3^{x^{2}-13 x+28}+\frac{2}{9}\right)=\log _{3} 3^{-1}
$$

Next we have

$$
\begin{aligned}
& 3^{x^{2}-13 x+28}+\frac{2}{9}=\frac{1}{3} ; 3^{x^{2}-13 x+28}=3^{-2} ; x^{2}-13 x+28=-2 \\
& x^{2}-13 x+30=0 ; x_{1}=3, x_{2}=10
\end{aligned}
$$

Answer: $x_{1}=3, x_{2}=10$.