7.166. $\sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\frac{1}{2} \log _{0.2} x+1 \geq 0, \\ \log _{0.2}+3 \geq 0, \quad \Leftrightarrow 00\end{array} \quad\right.$

Switch to base 0.2. We have

$$
\begin{aligned}
& \sqrt{\frac{1}{2} \log _{0.2} x+1}+\sqrt{\log _{0.2} x+3}=1 \Leftrightarrow \\
& \Leftrightarrow \sqrt{\log _{0.2} x+2}+\sqrt{2 \log _{0.2} x+6}=\sqrt{2}
\end{aligned}
$$

Squaring both sides of the equation, we get

$$
\begin{aligned}
& \log _{0.2} x+2+2 \sqrt{\left(\log _{0.2} x+2\right)\left(2 \log _{0.2} x+6\right)}+2 \log _{0.2} x+6=2 \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{\left(\log _{0.2} x+2\right)\left(2 \log _{0.2} x+6\right)}=-3 \log _{0.2} x-6 \Rightarrow \\
& \Rightarrow 4\left(\log _{0.2} x+2\right)\left(2 \log _{0.2} x+6\right)=9\left(\log _{0.2} x+2\right)^{2} \text { when }-3 \log _{0.2} x-6 \geq 0 \Leftrightarrow \\
& \Leftrightarrow \log _{0.2} x+2 \leq 0
\end{aligned}
$$

Considering the domain of definition, we have $\log _{0.2} x+2=0$, from which $x=25$.

Answer: 25.