4. Zadatak

Za sve $n\in\{1,2,\ldots,2019\}$ imamo

$$\begin{aligned} \log_{2^n}x\cdot\log_{2^{n+1}}x &=\frac{\log_2x}{\log_2 2^n}\cdot\frac{\log_2x}{\log_2 2^{n+1}}\\ &=\frac1{n(n+1)}(\log_2x)^2\\ &=\left(\frac1n-\frac1{n+1}\right)(\log_2x)^2, \end{aligned}$$

pa zadana jednadžba postaje

$$\left(1-\frac12+\frac12-\frac13+\cdots+\frac1{2019}-\frac1{2020}\right)(\log_2x)^2 =\frac{2019}{2020},$$

odnosno

$$\left(1-\frac1{2020}\right)(\log_2x)^2=\frac{2019}{2020},$$

odakle dobivamo $(\log_2x)^2=1$.

Odavde slijedi $\log_2x=1$ ili $\log_2x=-1$ pa su sva realna rješenja zadane jednadžbe

$$x_1=2\qquad\text{i}\qquad x_2=\frac12.$$