Exercise
Solve the inequality:
\log_2 (x^2-2x)-3\leq 0
Final Answer
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<x\leq 4
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