2. Zadatak

Da bi nejednadžba imala smisla mora biti zadovoljeno

$$x>0,\quad x\neq1\quad\text{i}\quad x\neq\frac12\tag{1}$$

Sada imamo:

$$\frac1{\log_2x}\cdot\frac1{\log_2(2x)}\cdot\log_2(16x)<1,$$ $$\frac1{\log_2x}\cdot\frac1{1+\log_2x}\cdot(4+\log_2x)<1.$$

Uz supstituciju

$$\log_2x=t,\tag{2}$$

dobivamo

$$\begin{aligned}\frac{4+t}{t(1+t)}&<1\\\frac{4-t^2}{t(1+t)}&<0.\end{aligned}$$

Rješenje ove nejednadžbe je

$$t\in\langle-\infty,-2\rangle\cup\langle-1,0\rangle\cup\langle2,\infty\rangle.\tag{3}$$

Iz (1), (2) i (3) dobivamo $x\in\left\langle0,\dfrac14\right\rangle\cup\left\langle\dfrac12,1\right\rangle\cup\langle4,\infty\rangle$.