Mister Exam
(4*x-7)*log(x^2-4*x+5)(3*x-5)>0 inequation
The solution
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/ 2 \ (4*x - 7)*log\x - 4*x + 5/*(3*x - 5) > 0
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) > 0$$
((4*x - 7)*log(x^2 - 4*x + 5))*(3*x - 5) > 0
Detail solution
Given the inequality:
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) = 0$$
Solve:
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
This roots
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{5}{3}$$
=
$$\frac{47}{30}$$
substitute to the expression
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) > 0$$
$$\left(-7 + \frac{4 \cdot 47}{30}\right) \log{\left(\left(- \frac{4 \cdot 47}{30} + \left(\frac{47}{30}\right)^{2}\right) + 5 \right)} \left(-5 + \frac{3 \cdot 47}{30}\right) > 0$$
one of the solutions of our inequality is:
$$x < \frac{5}{3}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{5}{3}$$
$$x > \frac{7}{4} \wedge x < 2$$
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) = 0$$
Solve:
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
This roots
$$x_{1} = \frac{5}{3}$$
$$x_{2} = \frac{7}{4}$$
$$x_{3} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{5}{3}$$
=
$$\frac{47}{30}$$
substitute to the expression
$$\left(4 x - 7\right) \log{\left(\left(x^{2} - 4 x\right) + 5 \right)} \left(3 x - 5\right) > 0$$
$$\left(-7 + \frac{4 \cdot 47}{30}\right) \log{\left(\left(- \frac{4 \cdot 47}{30} + \left(\frac{47}{30}\right)^{2}\right) + 5 \right)} \left(-5 + \frac{3 \cdot 47}{30}\right) > 0$$
/1069\
11*log|----|
\900 / > 0
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50 one of the solutions of our inequality is:
$$x < \frac{5}{3}$$
_____ _____
\ / \
-------ο-------ο-------ο-------
x1 x2 x3Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{5}{3}$$
$$x > \frac{7}{4} \wedge x < 2$$
Solving inequality on a graph
Rapid solution 2
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(-oo, 5/3) U (7/4, 2) U (2, oo)
$$x\ in\ \left(-\infty, \frac{5}{3}\right) \cup \left(\frac{7}{4}, 2\right) \cup \left(2, \infty\right)$$
x in Union(Interval.open(-oo, 5/3), Interval.open(7/4, 2), Interval.open(2, oo))
Rapid solution
[src]
Or(And(7/4 < x, x < 2), And(2 < x, x < oo), x < 5/3)
$$\left(\frac{7}{4} < x \wedge x < 2\right) \vee \left(2 < x \wedge x < \infty\right) \vee x < \frac{5}{3}$$
(x < 5/3)∨((7/4 < x)∧(x < 2))∨((2 < x)∧(x < oo))