2. Zadatak

Nejednadžba ima smisla ako je $x>0$ i $x\ne1$. Primijenimo svojstva logaritma potencije na lijevu stranu nejednadžbe.

$$\frac{1}{\log_x2\cdot2\log_x2}+\frac{1}{2\log_x2\cdot3\log_x2}+\cdots+\frac{1}{2013\log_x2\cdot2014\log_x2}\le2013\log_2x$$ $$\frac{1}{2(\log_x2)^2}+\frac{1}{2\cdot3(\log_x2)^2}+\cdots+\frac{1}{2013\cdot2014(\log_x2)^2}\le2013\log_2x.$$

Kako je

$$\frac{1}{\log_x2}=\log_2x$$

slijedi

$$\frac{(\log_2x)^2}{2}+\frac{(\log_2x)^2}{2\cdot3}+\cdots+\frac{(\log_2x)^2}{2013\cdot2014}\le2013\log_2x,$$ $$(\log_2x)^2\left(\frac12+\frac{1}{2\cdot3}+\cdots+\frac{1}{2013\cdot2014}\right)\le2013\log_2x,$$ $$(\log_2x)^2\left(1-\frac12+\frac12-\frac13+\cdots+\frac1{2013}-\frac1{2014}\right)\le2013\log_2x,$$ $$(\log_2x)^2\left(1-\frac1{2014}\right)\le2013\log_2x,$$ $$(\log_2x)^2\cdot\frac{2013}{2014}\le2013\log_2x,$$ $$(\log_2x)^2-2014\log_2x\le0,$$ $$\log_2x\cdot(\log_2x-2014)\le0.$$

Slijedi $\log_2x\ge0$ i $\log_2x\le2014$, odnosno $1\le x\le2^{2014}$. Zbog uvjeta s početka, rješenje je

$$x\in\langle1,2^{2014}].$$