fbpx
calculus online - Free exercises and solutions to help you succeed!

Inequalities – Inequality with log – Exercise 5714

Exercise

Solve the inequality:

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

Final Answer

-2\leq x\leq 0 \text{  or  } 2<x\leq 4

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

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply