## Task B-4.7.

Determine all natural numbers $x$ that are solutions to the inequality

$$
\log _{x}^{4} 2017+6 \cdot \log _{x}^{2} 2017>4 \cdot \log _{x}^{3} 2017+4 \cdot \log _{x} 2017
$$
## Second Solution.

The condition for solving the problem is $x \neq 1$.

Let's introduce the substitution $t=\log _{x} 2017$.

Notice that the number $t$ is always positive because $x$ is a natural number greater than 1. (The logarithm of a number greater than 1, with a base greater than 1, is always positive.)

The initial inequality transforms into

$$
\begin{aligned}
& t^{4}+6 t^{2}>4 t^{3}+4 t \\
& \Leftrightarrow t^{4}-4 t^{3}+6 t^{2}-4 t>0 \\
& \Leftrightarrow t\left(t^{3}-4 t^{2}+6 t-4\right)>0 \\
& \Leftrightarrow t\left(t^{3}-2 t^{2}-2 t^{2}+4 t+2 t-4\right)>0 \\
& \Leftrightarrow t\left(t^{2}(t-2)-2 t(t-2)+2(t-2)\right)>0 \\
& \Leftrightarrow t(t-2)\left(t^{2}-2 t+2\right)>0
\end{aligned}
$$

The trinomial $t^{2}-2 t+2$ has a positive value for all real numbers $t$. We have already noted that the number $t$ is always positive. Therefore, the last inequality is equivalent to

$$
t-2>0 \Leftrightarrow t>2
$$

Then $\log _{x} 2017>2$, which, under the condition $x>1$, gives $2017>x^{2}$.

It follows that $|x|1$ gives

$$
x2$ is worth 6 points, and its solution and the final solution of the problem are worth 4 points.