2. Determine the maximum value of the expression

$$
\log _{2}^{4} x+12 \log _{2}^{2} x \cdot \log _{2}\left(\frac{8}{x}\right)
$$

if the variable $x$ varies on the interval $[1,64]$.
Solution. Let $\log _{2} x=1$. We observe that $t$ increases from 0 to 6 as $x$ increases from 1 to 64. Further,

$$
\begin{aligned}
\log _{2}^{4} x+12 \log _{2}^{2} x \cdot \log _{2}\left(\frac{8}{x}\right) & =\log _{2}^{4} x+12 \log _{2}^{2} x \cdot\left(3-\log _{2} x\right) \\
& =t^{4}-12 t^{3}+36 t^{2}=(t(t-6))^{2}
\end{aligned}
$$

The function $f(t)=t(t-6)$ takes negative values for $0<t<6$ and reaches its minimum at the point $t=3$. Therefore, the maximum of the given expression is equal to $(f(3))^{2}=81$.