Mister Exam
lg^2x-4lgx-5>0 inequation
The solution
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[src]
2 log (x) - 4*log(x) - 5 > 0
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 > 0$$
log(x)^2 - 4*log(x) - 5 > 0
Detail solution
Given the inequality:
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 = 0$$
Solve:
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
This roots
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{-1}$$
=
$$- \frac{1}{10} + e^{-1}$$
substitute to the expression
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 > 0$$
$$-5 + \left(\log{\left(- \frac{1}{10} + e^{-1} \right)}^{2} - 4 \log{\left(- \frac{1}{10} + e^{-1} \right)}\right) > 0$$
one of the solutions of our inequality is:
$$x < e^{-1}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < e^{-1}$$
$$x > e^{5}$$
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 = 0$$
Solve:
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
This roots
$$x_{1} = e^{-1}$$
$$x_{2} = e^{5}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{-1}$$
=
$$- \frac{1}{10} + e^{-1}$$
substitute to the expression
$$\left(\log{\left(x \right)}^{2} - 4 \log{\left(x \right)}\right) - 5 > 0$$
$$-5 + \left(\log{\left(- \frac{1}{10} + e^{-1} \right)}^{2} - 4 \log{\left(- \frac{1}{10} + e^{-1} \right)}\right) > 0$$
2/ 1 -1\ / 1 -1\
-5 + log |- -- + e | - 4*log|- -- + e | > 0
\ 10 / \ 10 / one of the solutions of our inequality is:
$$x < e^{-1}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < e^{-1}$$
$$x > e^{5}$$
Solving inequality on a graph
Rapid solution 2
[src]
-1 5 (0, e ) U (e , oo)
$$x\ in\ \left(0, e^{-1}\right) \cup \left(e^{5}, \infty\right)$$
x in Union(Interval.open(0, exp(-1)), Interval.open(exp(5), oo))
Rapid solution
[src]
/ / -1\ / 5 \\ Or\And\0 < x, x < e /, And\x < oo, e < x//
$$\left(0 < x \wedge x < e^{-1}\right) \vee \left(x < \infty \wedge e^{5} < x\right)$$
((0 < x)∧(x < exp(-1)))∨((x < oo)∧(exp(5) < x))