3 . If $\mathrm{iog}_{2}\left(\log _{8} x\right)=\log _{8}\left(\log _{2} x\right)$, find $\left(\log _{2} x\right)^{2}$.
$$
\begin{array}{l}
3 .(27) \\
\because \log _{8} x=\frac{1}{\log _{x} 3}=\frac{1}{3 \log _{x} 2}=\frac{1}{3} \log _{2} x, \\
\log _{8}\left(\log _{2} x\right)=\frac{1}{3} \log _{2}\left(\log _{2} x\right),
\end{array}
$$

Let $y=\log _{2} x$, then the equation in the problem is equivalent to
$$
\log _{2} \frac{y}{3}=\frac{1}{3} \log _{2} y .
$$

From (1), we get
$$
\begin{aligned}
& \log _{2}\left(\frac{y}{3}\right)^{3}=\log _{2} y, \\
\therefore \quad & \left(\frac{y}{3}\right)^{3}=y .
\end{aligned}
$$

Simplifying, we get $y\left(y^{2}-27\right)=0$, ( 2 ) $\therefore y^{2}=\left(\log _{2} x\right)^{2}=27$.