Sol. We rewrite the inequality in the form

log |x| ( – x –1) ≥ log |x| (|x|)

This inequality is equivalent to the collection of systems

,

i.e. to the collection

, …(1)

The first system of collection (1) is equivalent to the collection of systems

, …(2)

We have,

⇒ , ⇒

⇒ ,

Since < 1, the first system of collection (2) has no solutions.

Let us solve the second system of collection (2).

We have

⇒ ⇒

⇒ – 2 ≤ x < – 1

Thus the interval –2 ≤ x < –1 is the set of all solutions of collection (2), i.e. of the first system of collection (1).

The second system of collection (1) is equivalent to the collection of the system.

…(3)

We solve the first system of collection (3), we have

⇒

⇒

The intervals – ∞ < x ≤ and < x < ∞ are the solution of the inequality 5x 2 + 4x – 8 ≥ 0, and the interval < x < is the solution of the inequality 2x 2 + 2x – 8 < 0.

Since < 1 < , it follows that

⇒ ≤ x < 1

i.e. the interval ≤ x < 1 is the solution of the first system of collection (3).

The second system of collection (3) has no solutions

Since ⇒ ⇒

and the last system has no solutions.

Thus the interval ≤ x < 1 is the set of all solutions of collection (3), i.e. of the second system of collection (1).

Thus the set of all solutions of collection (1) and consequently, of the original inequality consists of the intervals.– 2 ≤ x < – 1, ≤ x < 1