7.080. $\lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}6+x \geq 0, \\ x>0, \\ x \neq 1,\end{array} \quad 0<x \neq 1\right.$.

We will switch to base 10. We have

$\lg (\sqrt{6+x}+6)=2 \lg \sqrt{x}, \quad \lg (\sqrt{6+x}+6)=\lg x$.

Then $\sqrt{6+x}+6=x, \sqrt{6+x}=x-6 \Rightarrow\left\{\begin{array}{l}x^{2}-13 x+30=0, \\ x \geq 6,\end{array}\right.$

from which $x=10$.

Answer: 10.