$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$. 

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$7.25 \lg (\sqrt{6+x}+6)=\frac{2}{\log _{\sqrt{x}} 10}$.
7.25 From the definition of the logarithm, it follows that $\sqrt{x}>0, \sqrt{x} \neq 1$, i.e., $x>0, x \neq 1$. Applying formulas (7.8) and (7.6), we get $\lg (\sqrt{6+x}+6)=\lg x$. According to the hint $4^{0}$, we transition to an equivalent system of equations:

$\left\{\begin{array}{l}x>0, x \neq 1, \\ \sqrt{6+x}+6=x,\end{array}\right.$ or $\left\{\begin{array}{l}x>0, x \neq 1, \\ \sqrt{6+x}=x-6 .\end{array}\right.$

In the left part of the equation $\sqrt{6+x} \geq 0$, therefore $x-6 \geq 0$, i.e., $x \geq 6$, and both sides of the equation can be squared. We have

$$
\left\{\begin{array} { l } 
{ x \geq 6 } \\
{ 6 + x = x ^ { 2 } - 1 2 x + 3 6 , }
\end{array} \text { or } \left\{\begin{array}{l}
x \geq 6 \\
x^{2}-13 x+30=0
\end{array}\right.\right.
$$

We find the roots of the equation: $x_{1}=3, x_{2}=10$; of these, only $x=10$ satisfies the inequality $x \geq 6$.

Answer: $x=10$.