$7.43 \quad 5^{x} \sqrt[x]{8^{x-1}}=500$.

$7.44 \quad 5^{\frac{x}{\sqrt{x}+2}} \cdot 0.2^{\frac{4}{\sqrt{x}+2}}=125^{x-4} \cdot 0.04^{x-2}$.
7.43 According to the definition of the root, $x \neq 0$. Let's write the equation as

$$
5^{x} \cdot 2^{\frac{3 x-3}{x}}=5^{3} \cdot 2^{2}
$$

Dividing both sides by $5^{3} \cdot 2^{2} \neq 0$, we get

$$
5^{x-3} \cdot 2^{\frac{3 x-3}{x}-2}=1, \text { or } 5^{x-3} \cdot 2^{\frac{x-3}{x}}=1, \text { or }\left(5 \cdot 2^{\frac{1}{x}}\right)^{x-3}=1
$$

Since $5 \cdot 2^{\frac{1}{x}}>0$ and $5 \cdot 2^{\frac{1}{x}} \neq 1$, then $x-3=0$, i.e., $x=3$.

Answer: $x=3$.