7.066. $2^{\frac{1}{\sqrt{x}-1}} \cdot 0.5^{\frac{1}{\sqrt{x}+1}}=4^{\frac{\sqrt{x}}{x+\sqrt{x}}}$
Solution.

Domain of definition: $0<x \neq 1$.

We have: $2^{\frac{1}{\sqrt{x}-1}} \cdot 2^{-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}, 2^{\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}}=2^{\frac{2 \sqrt{x}}{x+\sqrt{x}}}$.

Then $\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}=\frac{2 \sqrt{x}}{x+\sqrt{x}}, x-\sqrt{x}-2=0$. Solving this equation as a quadratic equation in terms of $\sqrt{x}$, we find $\sqrt{x}=-1, \varnothing$; or $\sqrt{x}=2$, from which we have $x=4$.

Answer: 4.