Task B-3.2. Determine all real numbers $x$ for which

$$
\frac{1}{\log _{x} 2 \cdot \log _{x} 4}+\frac{1}{\log _{x} 4 \cdot \log _{x} 8}+\ldots+\frac{1}{\log _{x} 2^{2013} \cdot \log _{x} 2^{2014}} \leq 2013 \log _{2} x
$$
## Solution.

The inequality makes sense if $x>0$ and $x \neq 1$. Apply the properties of the logarithm of a power to the left side of the inequality.

$$
\begin{aligned}
& \frac{1}{\log _{x} 2 \cdot 2 \log _{x} 2}+\frac{1}{2 \log _{x} 2 \cdot 3 \log _{x} 2}+\ldots+\frac{1}{2013 \log _{x} 2 \cdot 2014 \log _{x} 2} \leq 2013 \log _{2} x \\
& \frac{1}{2\left(\log _{x} 2\right)^{2}}+\frac{1}{2 \cdot 3\left(\log _{x} 2\right)^{2}}+\ldots+\frac{1}{2013 \cdot 2014\left(\log _{x} 2\right)^{2}} \leq 2013 \log _{2} x
\end{aligned}
$$

Since $\frac{1}{\log _{x} 2}=\log _{2} x$, it follows that

$$
\begin{aligned}
& \frac{\left(\log _{2} x\right)^{2}}{2}+\frac{\left(\log _{2} x\right)^{2}}{2 \cdot 3}+\ldots+\frac{\left(\log _{2} x\right)^{2}}{2013 \cdot 2014} \leq 2013 \log _{2} x \\
& \left(\log _{2} x\right)^{2} \cdot\left(\frac{1}{2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{2013 \cdot 2014}\right) \leq 2013 \log _{2} x \\
& \left(\log _{2} x\right)^{2} \cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{2013}-\frac{1}{2014}\right) \leq 2013 \log _{2} x \\
& \left(\log _{2} x\right)^{2} \cdot\left(1-\frac{1}{2014}\right) \leq 2013 \log _{2} x \\
& \left(\log _{2} x\right)^{2} \cdot \frac{2013}{2014} \leq 2013 \log _{2} x \\
& \left(\log _{2} x\right)^{2}-2014 \log _{2} x \leq 0 \\
& \log _{2} x \cdot\left(\log _{2} x-2014\right) \leq 0
\end{aligned}
$$

It follows that $\log _{2} x \geq 0$ and $\log _{2} x \leq 2014$, or $1 \leq x \leq 2^{2014}$.

Due to the initial conditions, the solution is $x \in\left\langle 1,2^{2014}\right]$.