1. If $\log _{2} \log _{8} x=\log _{8} \log _{2} x$, then $\log _{4} x=$ $\qquad$
2. 1. $\frac{3 \sqrt{3}}{2}$.
From the given, we know $\left(\log _{8} x\right)^{3}=\log _{2} x>0$.
Let $\log _{2} x=y$. Then $\left(\frac{y}{3}\right)^{3}=y \Rightarrow y=3 \sqrt{3}$.
Therefore, $\log _{4} x=\frac{3 \sqrt{3}}{2}$.