## Task B-3.6.

Solve the inequality $\log _{|x-1|} 2 x \geqslant 1$ in the set of real numbers.
## Solution.

Conditions: $2 x>0,|x-1|>0,|x-1| \neq 1$ so it must be $x>0, x \neq 1, x \neq 2$

1. case: $|x-1|>1$, i.e., $x>2$

$$
\begin{aligned}
\log _{|x-1|} 2 x & \geqslant 1 \\
2 x & \geqslant|x-1| \\
2 x & \geqslant x-1 \\
x & \geqslant-1
\end{aligned}
$$

Solution of the 1st case: $x \geqslant-1$ and $x>2$ gives $x>2$.

2. case: $0<|x-1|<1$ so it must be $0<x<2$ and $x \neq 1$.

$$
\begin{aligned}
\log _{|x-1|} 2 x & \geqslant 1 \\
2 x & \leqslant|x-1|
\end{aligned}
$$

I. If $0<x<1$ then $2 x \leqslant-x+1$, i.e., $3 x \leqslant 1$ From $x \leqslant \frac{1}{3}$ and $0<x<1$ it follows that $0<x \leqslant \frac{1}{3}$

II. If $1<x<2$ then $2 x \leqslant x-1$, i.e., $x \leqslant-1$.

1 point

$\mathrm{U}$ In this case, the intersection is an empty set.

2 points

Final solution: $x \in\left\langle 0, \frac{1}{3}\right] \cup\langle 2,+\infty\rangle$.

2 points