Mister Exam
ln(x^10)>5 inequation
The solution
Detail solution
Given the inequality:
$$\log{\left(x^{10} \right)} > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(x^{10} \right)} = 5$$
Solve:
$$x_{1} = \frac{\left(-1 + \sqrt{5} - \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{2} = \frac{\left(-1 + \sqrt{5} + \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{3} = \frac{\left(1 + \sqrt{5} - \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{4} = \frac{\left(1 + \sqrt{5} + \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{5} = - \frac{\left(1 + \sqrt{5} + \sqrt{2} i \sqrt{5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{6} = \frac{\left(- \sqrt{5} - 1 + \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{7} = \frac{\left(- \sqrt{5} + 1 - \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{8} = \frac{\left(- \sqrt{5} + 1 + \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{9} = - e^{\frac{1}{2}}$$
$$x_{10} = e^{\frac{1}{2}}$$
Exclude the complex solutions:
$$x_{1} = - e^{\frac{1}{2}}$$
$$x_{2} = e^{\frac{1}{2}}$$
This roots
$$x_{1} = - e^{\frac{1}{2}}$$
$$x_{2} = e^{\frac{1}{2}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- e^{\frac{1}{2}} - \frac{1}{10}$$
=
$$- e^{\frac{1}{2}} - \frac{1}{10}$$
substitute to the expression
$$\log{\left(x^{10} \right)} > 5$$
$$\log{\left(\left(- e^{\frac{1}{2}} - \frac{1}{10}\right)^{10} \right)} > 5$$
one of the solutions of our inequality is:
$$x < - e^{\frac{1}{2}}$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - e^{\frac{1}{2}}$$
$$x > e^{\frac{1}{2}}$$
$$\log{\left(x^{10} \right)} > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(x^{10} \right)} = 5$$
Solve:
$$x_{1} = \frac{\left(-1 + \sqrt{5} - \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{2} = \frac{\left(-1 + \sqrt{5} + \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{3} = \frac{\left(1 + \sqrt{5} - \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{4} = \frac{\left(1 + \sqrt{5} + \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{5} = - \frac{\left(1 + \sqrt{5} + \sqrt{2} i \sqrt{5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{6} = \frac{\left(- \sqrt{5} - 1 + \sqrt{2} \sqrt{-5 + \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{7} = \frac{\left(- \sqrt{5} + 1 - \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{8} = \frac{\left(- \sqrt{5} + 1 + \sqrt{2} \sqrt{-5 - \sqrt{5}}\right) e^{\frac{1}{2}}}{4}$$
$$x_{9} = - e^{\frac{1}{2}}$$
$$x_{10} = e^{\frac{1}{2}}$$
Exclude the complex solutions:
$$x_{1} = - e^{\frac{1}{2}}$$
$$x_{2} = e^{\frac{1}{2}}$$
This roots
$$x_{1} = - e^{\frac{1}{2}}$$
$$x_{2} = e^{\frac{1}{2}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- e^{\frac{1}{2}} - \frac{1}{10}$$
=
$$- e^{\frac{1}{2}} - \frac{1}{10}$$
substitute to the expression
$$\log{\left(x^{10} \right)} > 5$$
$$\log{\left(\left(- e^{\frac{1}{2}} - \frac{1}{10}\right)^{10} \right)} > 5$$
/ 10\
|/ 1 1/2\ |
log||- -- - e | | > 5
\\ 10 / /
one of the solutions of our inequality is:
$$x < - e^{\frac{1}{2}}$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - e^{\frac{1}{2}}$$
$$x > e^{\frac{1}{2}}$$
Solving inequality on a graph
Rapid solution
[src]
/ 1/2 1/2 \ Or\x < -e , e < x/
$$x < - e^{\frac{1}{2}} \vee e^{\frac{1}{2}} < x$$
(exp(1/2) < x)∨(x < -exp(1/2))
Rapid solution 2
[src]
1/2 1/2 (-oo, -e ) U (e , oo)
$$x\ in\ \left(-\infty, - e^{\frac{1}{2}}\right) \cup \left(e^{\frac{1}{2}}, \infty\right)$$
x in Union(Interval.open(-oo, -exp(1/2)), Interval.open(exp(1/2), oo))