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log2x(x+4)*logx(2-x)>0 inequation

A inequation with variable

The solution

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log(2*x)*(x + 4)*log(x)*(2 - x) > 0
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) > 0$$
(((x + 4)*log(2*x))*log(x))*(2 - x) > 0
Detail solution
Given the inequality:
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) > 0$$
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:
$$x_{0} < x_{1}$$
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
$$\left(x + 4\right) \log{\left(2 x \right)} \log{\left(x \right)} \left(2 - x\right) > 0$$
$$\left(- \frac{41}{10} + 4\right) \log{\left(\frac{\left(-41\right) 2}{10} \right)} \log{\left(- \frac{41}{10} \right)} \left(2 - - \frac{41}{10}\right) > 0$$
   /  log(41/5)   pi*I\ /          /41\\    
61*|- --------- - ----|*|pi*I + log|--||    
   \      10       10 / \          \10// > 0
----------------------------------------    
                   10                       

Then
$$x < -4$$
no execute
one of the solutions of our inequality is:
$$x > -4 \wedge x < \frac{1}{2}$$
         _____           _____  
        /     \         /     \  
-------ο-------ο-------ο-------ο-------
       x1      x2      x3      x4

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x > -4 \wedge x < \frac{1}{2}$$
$$x > 1 \wedge x < 2$$
Solving inequality on a graph
Rapid solution 2 [src]
(0, 1/2) U (1, 2)
$$x\ in\ \left(0, \frac{1}{2}\right) \cup \left(1, 2\right)$$
x in Union(Interval.open(0, 1/2), Interval.open(1, 2))
Rapid solution [src]
Or(And(0 < x, x < 1/2), And(1 < x, x < 2))
$$\left(0 < x \wedge x < \frac{1}{2}\right) \vee \left(1 < x \wedge x < 2\right)$$
((0 < x)∧(x < 1/2))∨((1 < x)∧(x < 2))