35. Find the maximum value of the expression $\log _{2}^{4} x+12 \log _{2}^{2} x \log _{2} \frac{8}{x}$, assuming that $x$ varies from 1 to 64.
35. Let's denote the given expression by $y$. Then it is easy to see that $y=\log _{2}^{2} x\left(6-\log _{2} x\right)^{2}$. But $1<x<64$, so $0<\log _{2} x<6$. The maximum value of $y$ coincides with the maximum value of the product. Since $\log _{2} x+(6-$ $\left.-\log _{2} x\right)=6$, the product will be the largest under the condition that

$\frac{\log _{2} x}{2}=\frac{6-\log _{2} x}{2}$; from which $\log _{2} x=3, x=8$. The maximum value of $y$ is 81.