2. If the real number $x$ satisfies $\log _{2} \log _{2} x=\log _{4} \log _{4} x$, then $x=$ $\qquad$ .
2. $\sqrt{2}$.

Let $y=\log _{a} \log _{a} x$. Then $a^{a^{y}}=x$. Let $a=2$ and $a=4$. According to the problem,
$$
\begin{array}{l}
2^{2^{y}}=4^{4^{y}}=\left(2^{2}\right)^{\left(2^{2}\right) y}=2^{2^{2 y+1}} \\
\Rightarrow y=2 y+1 \Rightarrow y=-1 \\
\Rightarrow x=\sqrt{2} .
\end{array}
$$