7.101. $\lg 8-\lg \sqrt{x+6}=\lg 16-\lg (x-2)$
Solution.

Domain of definition: $\left\{\begin{array}{l}x+6>0, \\ x-2>0,\end{array} \quad x>2\right.$.

We have

$$
\lg \frac{8}{\sqrt{x+6}}=\lg \frac{16}{x-2}, \quad \frac{8}{\sqrt{x+6}}=\frac{16}{x-2}, \quad 2 \sqrt{x+6}=x-2, x^{2}-8 x-20=0
$$

from which $x_{1}=10, x_{2}=-2 ; x_{2}=-2$ does not satisfy the domain of definition.

Answer: 10.