 # Inequalities – Inequality with log – Exercise 5714

Exercise

Solve the inequality:

$$\log_2 (x^2-2x)-3\leq 0$$

$$-2\leq x\leq 0 \text{ or } 2

Solution

$$\log_2 (x^2-2x)-3\leq 0$$

Move 3 to the other side:

$$\log_2 (x^2-2x)\leq 3$$

By logarithm definition we get:

$$x^2-2x\leq 2^3$$

Also, we require the phrase inside the log to be greater than zero:

$$x^2-2x>0$$

Because otherwise there is no solution to the inequality.

We solve the two inequalities we have received and intersect their results (“and”).

We solve the first inequality:

$$x^2-2x>0$$

$$x(x-2)>0$$

It is a square inequality. Its roots are 0 and 2.

The coefficient of the square expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. The solutions (= zeros = roots) of the quadratic equation are 0 and 2. Hence, the graph goes through the x-axis at these points. Therefore, the solution of the inequality is

$$x<0 \text{ or } x>2$$

We solve the second inequality:

$$x^2-2x\leq 2^3$$

We move everything to one side:

$$x^2-2x-8\leq 0$$

The coefficient of the square expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is below the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{2\pm \sqrt{{(-2)}^2-4\cdot 4\cdot (-8)}}{2\cdot 1}=$$

$$=\frac{2\pm \sqrt{36}}{2}=$$

$$=\frac{2\pm 6}{2}=$$

Hence, we get the solutions:

$$x_1=\frac{2+ 6}{2}=4$$

$$x_2=\frac{2- 6}{2}=-2$$

Because we are looking for the section below the x-axis or on it and the parabola “smiles”, we get that the solution of the inequality is

$$-2\leq x\leq 4$$

Finally, we intersect both results ( “and”) and get:

$$-2\leq x\leq 0 \text{ or } 2

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