Mister Exam
log2x(x+4)·logx(2-x)≤0 inequation
The solution
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log(2*x)*(x + 4)*log(x)*(2 - x) <= 0
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) \leq 0$$
(((x + 4)*log(2*x))*log(x))*(2 - x) <= 0
Detail solution
Given the inequality:
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) = 0$$
Solve:
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
This roots
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) \leq 0$$
$$\left(- \frac{41}{10} + 4\right) \log{\left(\frac{\left(-41\right) 2}{10} \right)} \log{\left(- \frac{41}{10} \right)} \left(2 - - \frac{41}{10}\right) \leq 0$$
Then
$$x \leq -4$$
no execute
one of the solutions of our inequality is:
$$x \geq -4 \wedge x \leq \frac{1}{2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq -4 \wedge x \leq \frac{1}{2}$$
$$x \geq 1 \wedge x \leq 2$$
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) = 0$$
Solve:
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
This roots
$$x_{1} = -4$$
$$x_{2} = \frac{1}{2}$$
$$x_{3} = 1$$
$$x_{4} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) \leq 0$$
$$\left(- \frac{41}{10} + 4\right) \log{\left(\frac{\left(-41\right) 2}{10} \right)} \log{\left(- \frac{41}{10} \right)} \left(2 - - \frac{41}{10}\right) \leq 0$$
/ log(41/5) pi*I\ / /41\\
61*|- --------- - ----|*|pi*I + log|--||
\ 10 10 / \ \10// <= 0
----------------------------------------
10 Then
$$x \leq -4$$
no execute
one of the solutions of our inequality is:
$$x \geq -4 \wedge x \leq \frac{1}{2}$$
_____ _____
/ \ / \
-------•-------•-------•-------•-------
x1 x2 x3 x4Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq -4 \wedge x \leq \frac{1}{2}$$
$$x \geq 1 \wedge x \leq 2$$
Solving inequality on a graph
Rapid solution
[src]
Or(And(1/2 <= x, x <= 1), 2 <= x)
$$\left(\frac{1}{2} \leq x \wedge x \leq 1\right) \vee 2 \leq x$$
(2 <= x)∨((1/2 <= x)∧(x <= 1))
Rapid solution 2
[src]
[1/2, 1] U [2, oo)
$$x\ in\ \left[\frac{1}{2}, 1\right] \cup \left[2, \infty\right)$$
x in Union(Interval(1/2, 1), Interval(2, oo))