Mister Exam
lg^2x+lgx-2>0 inequation
The solution
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[src]
2 log (x) + log(x) - 2 > 0
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 > 0$$
log(x)^2 + log(x) - 1*2 > 0
Detail solution
Given the inequality:
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 = 0$$
Solve:
$$x_{1} = e$$
$$x_{2} = e^{-2}$$
$$x_{1} = e$$
$$x_{2} = e^{-2}$$
This roots
$$x_{2} = e^{-2}$$
$$x_{1} = e$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{-2}$$
=
$$- \frac{1}{10} + e^{-2}$$
substitute to the expression
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 > 0$$
$$\log{\left(- \frac{1}{10} + e^{-2} \right)} - 2 + \log{\left(- \frac{1}{10} + e^{-2} \right)}^{2} > 0$$
one of the solutions of our inequality is:
$$x < e^{-2}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < e^{-2}$$
$$x > e$$
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 = 0$$
Solve:
$$x_{1} = e$$
$$x_{2} = e^{-2}$$
$$x_{1} = e$$
$$x_{2} = e^{-2}$$
This roots
$$x_{2} = e^{-2}$$
$$x_{1} = e$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{-2}$$
=
$$- \frac{1}{10} + e^{-2}$$
substitute to the expression
$$\log{\left(x \right)}^{2} + \log{\left(x \right)} - 2 > 0$$
$$\log{\left(- \frac{1}{10} + e^{-2} \right)} - 2 + \log{\left(- \frac{1}{10} + e^{-2} \right)}^{2} > 0$$
2/ 1 -2\ / 1 -2\
-2 + log |- -- + e | + log|- -- + e | > 0
\ 10 / \ 10 / one of the solutions of our inequality is:
$$x < e^{-2}$$
_____ _____
\ /
-------ο-------ο-------
x_2 x_1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < e^{-2}$$
$$x > e$$
Solving inequality on a graph
Rapid solution 2
[src]
-2 (0, e ) U (e, oo)
$$x\ in\ \left(0, e^{-2}\right) \cup \left(e, \infty\right)$$
x in Union(Interval.open(0, exp(-2)), Interval.open(E, oo))
Rapid solution
[src]
/ / -2\ \ Or\And\0 < x, x < e /, e < x/
$$\left(0 < x \wedge x < e^{-2}\right) \vee e < x$$
(E < x)∨((0 < x)∧(x < exp(-2)))