7.197. $5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$.
## Solution.

Domain of definition: $x \geq 0$.

From the condition we have

$$
\begin{aligned}
& 5^{\frac{x}{\sqrt{x}+2}} \cdot 5^{-\frac{4}{\sqrt{x}+2}}=5^{3 x-12} \cdot 5^{-2 x+4} \Leftrightarrow 5^{\frac{x}{\sqrt{x}+2}-\frac{4}{\sqrt{x}+2}}=5^{3 x-12-2 x+4} \Leftrightarrow \\
& \Leftrightarrow \frac{x-4}{\sqrt{x}+2}=x-8 \Leftrightarrow x \sqrt{x}+x-8 \sqrt{x}-12=0
\end{aligned}
$$

Let $\sqrt{x}=y \geq 0$. The equation in terms of $y$ becomes $y^{3}+y^{2}-8 y-12=0,(y-3)(y+2)^{2}=0$, from which $y_{1}=3, y_{2,3}=-2 ; y_{2,3}=-2$ does not fit. Then $\sqrt{x}=3, x=9$.

Answer: 9.