Mister Exam
(x^2e^x-4e^x+2x^2-8)log4(3–x)/log2^2(x-3)^2<=0 inequation
The solution
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/ 2 x x 2 \ log(3 - x)
\x *E - 4*E + 2*x - 8/*----------
log(4) 2
------------------------------------*(x - 3) <= 0
2
log (2)
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} \leq 0$$
(((log(3 - x)/log(4))*(2*x^2 + E^x*x^2 - 4*exp(x) - 8))/log(2)^2)*(x - 3)^2 <= 0
Detail solution
Given the inequality:
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} = 0$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = \log{\left(2 \right)} + i \pi$$
Exclude the complex solutions:
$$x_{1} = -2$$
$$x_{2} = 2$$
This roots
$$x_{1} = -2$$
$$x_{2} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} \leq 0$$
$$\frac{\frac{\log{\left(3 - - \frac{21}{10} \right)}}{\log{\left(4 \right)}} \left(-8 + \left(\left(- \frac{4}{e^{\frac{21}{10}}} + \frac{\left(- \frac{21}{10}\right)^{2}}{e^{\frac{21}{10}}}\right) + 2 \left(- \frac{21}{10}\right)^{2}\right)\right)}{\log{\left(2 \right)}^{2}} \left(-3 - \frac{21}{10}\right)^{2} \leq 0$$
but
Then
$$x \leq -2$$
no execute
one of the solutions of our inequality is:
$$x \geq -2 \wedge x \leq 2$$
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} = 0$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = \log{\left(2 \right)} + i \pi$$
Exclude the complex solutions:
$$x_{1} = -2$$
$$x_{2} = 2$$
This roots
$$x_{1} = -2$$
$$x_{2} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\frac{\frac{\log{\left(3 - x \right)}}{\log{\left(4 \right)}} \left(\left(2 x^{2} + \left(e^{x} x^{2} - 4 e^{x}\right)\right) - 8\right)}{\log{\left(2 \right)}^{2}} \left(x - 3\right)^{2} \leq 0$$
$$\frac{\frac{\log{\left(3 - - \frac{21}{10} \right)}}{\log{\left(4 \right)}} \left(-8 + \left(\left(- \frac{4}{e^{\frac{21}{10}}} + \frac{\left(- \frac{21}{10}\right)^{2}}{e^{\frac{21}{10}}}\right) + 2 \left(- \frac{21}{10}\right)^{2}\right)\right)}{\log{\left(2 \right)}^{2}} \left(-3 - \frac{21}{10}\right)^{2} \leq 0$$
/ -21 \
| ----|
| 10 |
|41 41*e | /51\
2601*|-- + --------|*log|--| <= 0
\50 100 / \10/
----------------------------
2
100*log (2)*log(4) but
/ -21 \
| ----|
| 10 |
|41 41*e | /51\
2601*|-- + --------|*log|--| >= 0
\50 100 / \10/
----------------------------
2
100*log (2)*log(4) Then
$$x \leq -2$$
no execute
one of the solutions of our inequality is:
$$x \geq -2 \wedge x \leq 2$$
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