Mister Exam
(1/3)^(x/2)>9 inequation
The solution
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-x --- 2 3 > 9
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} > 9$$
(1/3)^(x/2) > 9
Detail solution
Given the inequality:
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} > 9$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} = 9$$
Solve:
Given the equation:
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} = 9$$
or
$$-9 + \left(\frac{1}{3}\right)^{\frac{x}{2}} = 0$$
or
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = 9$$
or
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = 9$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{\sqrt{3}}{3}\right)^{x}$$
we get
$$v - 9 = 0$$
or
$$v - 9 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 9$$
do backward replacement
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(\frac{\sqrt{3}}{3} \right)}}$$
$$x_{1} = 9$$
$$x_{1} = 9$$
This roots
$$x_{1} = 9$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 9$$
=
$$\frac{89}{10}$$
substitute to the expression
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} > 9$$
$$\left(\frac{1}{3}\right)^{\frac{89}{2 \cdot 10}} > 9$$
Then
$$x < 9$$
no execute
the solution of our inequality is:
$$x > 9$$
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} > 9$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} = 9$$
Solve:
Given the equation:
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} = 9$$
or
$$-9 + \left(\frac{1}{3}\right)^{\frac{x}{2}} = 0$$
or
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = 9$$
or
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = 9$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{\sqrt{3}}{3}\right)^{x}$$
we get
$$v - 9 = 0$$
or
$$v - 9 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 9$$
do backward replacement
$$\left(\frac{\sqrt{3}}{3}\right)^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(\frac{\sqrt{3}}{3} \right)}}$$
$$x_{1} = 9$$
$$x_{1} = 9$$
This roots
$$x_{1} = 9$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 9$$
=
$$\frac{89}{10}$$
substitute to the expression
$$\left(\frac{1}{3}\right)^{\frac{x}{2}} > 9$$
$$\left(\frac{1}{3}\right)^{\frac{89}{2 \cdot 10}} > 9$$
11
--
20
3 > 9
---
243
Then
$$x < 9$$
no execute
the solution of our inequality is:
$$x > 9$$
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x1
Solving inequality on a graph
Rapid solution
[src]
-2*log(9)
x < ---------
log(3)
$$x < - \frac{2 \log{\left(9 \right)}}{\log{\left(3 \right)}}$$
x < -2*log(9)/log(3)
Rapid solution 2
[src]
-2*log(9)
(-oo, ---------)
log(3)
$$x\ in\ \left(-\infty, - \frac{2 \log{\left(9 \right)}}{\log{\left(3 \right)}}\right)$$
x in Interval.open(-oo, -2*log(9)/log(3))