7.102. $2 \lg \sqrt{4-x}+\lg (6-x)=1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}4-x>0, \\ 6-x>0,\end{array} \quad x<4\right.$.

Rewrite the equation as

$$
\lg (4-x)+\lg (6-x)=1, \quad \lg (4-x)(6-x)=1
$$

from which $(4-x)(6-x)=10, x^{2}+10 x-14=0$. Therefore, $x_{1}=5-\sqrt{11}$, $x_{2}=5+\sqrt{11} ; x_{2}=5+\sqrt{11}$ does not satisfy the domain of definition.

Answer: $\quad 5-\sqrt{11}$.