7.069. $\log _{\sqrt{5}}\left(4^{x}-6\right)-\log _{\sqrt{5}}\left(2^{x}-2\right)=2$.
Solution.

Domain of Definition (DOD): $\left\{\begin{array}{l}4^{x}-6>0 \\ 2^{x}-2>0 .\end{array}\right.$

We have $\log _{\sqrt{5}} \frac{4^{x}-6}{2^{x}-2}=2, \frac{2^{2 x}-6}{2^{2}-2}=5,2^{2 x}-5 \cdot 2^{x}+4=0$. Solving this equation as a quadratic in terms of $2^{x}$, we find $\left(2^{x}\right)=1$, from which we get $x_{1}=0$, or $\left(2^{x}\right)_{2}=4$, from which we get $x_{2}=2 ; x_{1}=0$ does not satisfy the DOD.

Answer: 2.