6. If $a>0, b>0$, and $\arcsin a+\arcsin b=\frac{\pi}{2}, m=\log _{2} \frac{1}{a}+\log _{2} \frac{1}{b}$, then the range of $m$ is $\qquad$ .
6. $[1,+\infty)$.

Let $\arcsin a=\alpha, \arcsin b=\beta$, then $a=\sin \alpha, b=\sin \beta, \alpha+\beta=\frac{\pi}{2}$, so $a^{2}+b^{2}=1$. $a^{2}+b^{2} \geqslant 2 a b \Rightarrow a b \leqslant \frac{a^{2}+b^{2}}{2}=\frac{1}{2} \Rightarrow \frac{1}{a b} \geqslant 2$, thus $\log _{2} \frac{1}{a}+\log _{2} \frac{1}{b}=\log _{2} \frac{1}{a b} \geqslant \log _{2} \frac{1}{\frac{1}{2}}=1$. Therefore, $\log _{2} \frac{1}{a}+\log _{2} \frac{1}{b} \in[1,+\infty)$.