$7.73 \sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$.

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$7.73 \sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$.
7.73 Given the domains of the logarithmic function and the square root, we obtain the system of inequalities

$$
\left\{\begin{array}{l}
x>0 \\
\log _{0.04} x+1 \geq 0 \\
\log _{0.2} x+3 \geq 0
\end{array}\right.
$$

We use the fact that $\log _{0.2} x=\log _{0.04} x^{2}=2 \log _{0.04} x$ (since $x>0$). Then, setting $\log _{0.04} x=y$, we get the equation

$$
\sqrt{y+1}+\sqrt{2 y+3}=1, \text { or } \sqrt{2 y+3}=1-\sqrt{y+1} \text {. }
$$

Squaring both sides of the equation, we get:

$2 y+3=1+y+1-2 \sqrt{y+1}$, or $2 \sqrt{y+1}=-1-y$.

Since $2 \sqrt{y+1} \geq 0$, then $-1-y \geq 0$, from which $y \leq-1$. Squaring both sides of the equation again, we have $4(y+1)=1+y^{2}+2 y$, or $y^{2}-2 y-3=0$, from which $y_{1}=-1, \quad y_{2}=3$ (does not satisfy the condition $y \leq-1$). Therefore, $\log _{0.04} x=-1$, i.e., $x=0.04^{-1}=25$. In this case, the conditions (1) are satisfied.

Answer: $x=25$.