## Task B-3.1.

Determine the maximum value of the function $f(x)=\log _{2}^{4} x+12 \log _{2}^{2} x \cdot \log _{2} \frac{8}{x}$ on the interval $[1,64]$.
## Solution.

By using the properties of logarithms, the function $f(x)=\log _{2}^{4} x+12 \log _{2}^{2} x \cdot \log _{2} \frac{8}{x}$ can be written in the form

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

From $x \in[1,64]$ and $\log _{2} x$ being an increasing function, it follows that $\log _{2} x \in\left[\log _{2} 1, \log _{2} 64\right]=[0,6]$. Therefore, it is sufficient to consider the function $g(t)=t^{2}-6 t$ on the interval $[0,6]$.

The function $g$ is a quadratic function that achieves its minimum value of -9 at $t=3$.

Notice that 0 and 6 are the roots of the function $g$. Therefore, on the interval $[0,6]$, the function $g$ takes all values in the interval $[-9,0]$. For these values of $g$, the function $g^{2}$ takes all values in the interval $[0,81]$ and its maximum value is 81, which it achieves for $g(t)=-9$, i.e., for $t=3$.

We conclude that the function $f=g^{2}$ on the interval [1,64] has the maximum value 81, which it achieves for $x$ such that $\log _{2} x=3$, i.e., for $x=8$.