7.024. $\lg 5+\lg (x+10)=1-\lg (2 x-1)+\lg (21 x-20)$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+10>0, \\ 2 x-1>0, \\ 21 x-20>0,\end{array} \quad x>\frac{20}{21}\right.$.

$$
\begin{aligned}
& \lg 5+\lg (x+10)=\lg 10-\lg (2 x-1)+\lg (21 x-20) \Leftrightarrow \lg 5(x+10)= \\
& =\lg \frac{10 \cdot(21 x-20)}{2 x-1} \Rightarrow 5(x+10)=\frac{10 \cdot(21 x-20)}{2 x-1}
\end{aligned}
$$

from which $2 x^{2}-23 x+30=0$. Solving this equation, we have $x_{1}=1.5 ; x_{2}=10$.

Answer: 1.5 ; 10.