152 If $x$ varies in the solution set obtained in the previous problem, then the maximum value of $f(x)=\left(\log _{2} \frac{x}{3}\right)\left(\log _{2} \frac{x}{4}\right)$ is equal to
$$
1523-\log _{2} 3. \text{ For the } x \text{ in the above problem, we have } \frac{3}{2} \leqslant \log _{2} x \leqslant 3. \text{ Let } \log _{2} x=u, \text{ then }
$$
$$
\begin{array}{c}
f(x)=\left(\log _{2} \frac{x}{3}\right)\left(\log _{2} \frac{x}{4}\right)=\left(u-\log _{2} 3\right)(u-2) \\
=u^{2}-\left(2+\log _{2} 3\right) u+2 \log _{2} 3 \\
\text { Also, } \frac{2+\log _{2} 3}{2}=\log _{2} \sqrt{12} \text {, and } \frac{3}{2}<\log _{2} \sqrt{12}<3, \text { and } \\
\log _{2} \sqrt{12}-\frac{1}{2}\left(\frac{3}{2}+3\right)=\log _{2} \sqrt{12}-\frac{9}{4}=\log _{2} \frac{\sqrt{3}}{2 \sqrt[4]{2}}<0 .
\end{array}
$$

It is clear that the axis of symmetry of the parabola $u=\log _{2} \sqrt{12}$ is closer to $u=\frac{3}{2}$, hence $f(x)$ reaches its maximum value $3-\log _{2} 3$ when $u=3$, i.e., $x=8$.