7. $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is

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$$
\log _{4}(x+2 y)+\log _{4}(x-2 y)=\log _{4}\left(x^{2}-4 y^{2}\right)=1 \Rightarrow x^{2}-4 y^{2}=4 \text {. Let } x=2 \sec \theta, y=\tan \theta \text {, }
$$

then $|x|-|y|=2 \sec \theta-\tan \theta=\frac{2-\sin \theta}{\cos \theta} \geqslant \sqrt{3}$, equality holds when $\theta=\frac{\pi}{6}$. Therefore, the minimum value of $|x|-|y|$ is $\sqrt{3}$.