7.103. $\frac{\lg (2 x-19)-\lg (3 x-20)}{\lg x}=-1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}2 x-19>0, \\ 3 x-20>0,\end{array} \quad x>\frac{19}{2}\right.$.

## From the condition

$$
\begin{aligned}
& \lg (2 x-19)-\lg (3 x-20)=-\lg x, \lg (2 x-19)+\lg x=\lg (3 x-20) \\
& x(2 x-19)=3 x-20, x^{2}-11 x+10=0 .
\end{aligned}
$$

From here $x_{1}=10, x_{2}=1 ; x_{2}=1$ does not satisfy the domain of definition.

Answer: 10.