Mister Exam
Other calculators
- Square Inequality Step-by-Step
- Inequality with the module Step by Step
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-
How to use it?
- Inequation:
- (abs((x+2)/(x-6)))<=(abs((x+2)/(x-3)))
- cos3x>=sqrt-2/2
- (1/2)^(x+2)>4
-
sin×(7/15)×x>=sqrt(3)/2
-
Identical expressions
- (one / two)^(x+ two)> four
- (1 divide by 2) to the power of (x plus 2) greater than 4
- (one divide by two) to the power of (x plus two) greater than four
- (1/2)(x+2)>4
- 1/2x+2>4
- 1/2^x+2>4
- (1 divide by 2)^(x+2)>4
-
Similar expressions
- (1/2)^(x-2)>4
(1/2)^(x+2)>4 inequation
The solution
Detail solution
Given the inequality:
$$\left(\frac{1}{2}\right)^{x + 2} > 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{2}\right)^{x + 2} = 4$$
Solve:
Given the equation:
$$\left(\frac{1}{2}\right)^{x + 2} = 4$$
or
$$\left(\frac{1}{2}\right)^{x + 2} - 4 = 0$$
or
$$\frac{2^{- x}}{4} = 4$$
or
$$\left(\frac{1}{2}\right)^{x} = 16$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{2}\right)^{x}$$
we get
$$v - 16 = 0$$
or
$$v - 16 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 16$$
do backward replacement
$$\left(\frac{1}{2}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = 16$$
$$x_{1} = 16$$
This roots
$$x_{1} = 16$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 16$$
=
$$\frac{159}{10}$$
substitute to the expression
$$\left(\frac{1}{2}\right)^{x + 2} > 4$$
$$\left(\frac{1}{2}\right)^{2 + \frac{159}{10}} > 4$$
Then
$$x < 16$$
no execute
the solution of our inequality is:
$$x > 16$$
$$\left(\frac{1}{2}\right)^{x + 2} > 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{2}\right)^{x + 2} = 4$$
Solve:
Given the equation:
$$\left(\frac{1}{2}\right)^{x + 2} = 4$$
or
$$\left(\frac{1}{2}\right)^{x + 2} - 4 = 0$$
or
$$\frac{2^{- x}}{4} = 4$$
or
$$\left(\frac{1}{2}\right)^{x} = 16$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{2}\right)^{x}$$
we get
$$v - 16 = 0$$
or
$$v - 16 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 16$$
do backward replacement
$$\left(\frac{1}{2}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = 16$$
$$x_{1} = 16$$
This roots
$$x_{1} = 16$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 16$$
=
$$\frac{159}{10}$$
substitute to the expression
$$\left(\frac{1}{2}\right)^{x + 2} > 4$$
$$\left(\frac{1}{2}\right)^{2 + \frac{159}{10}} > 4$$
10___
\/ 2
------ > 4
262144
Then
$$x < 16$$
no execute
the solution of our inequality is:
$$x > 16$$
_____
/
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x1
Solving inequality on a graph
Rapid solution
[src]
log(4)
x < -2 - ------
log(2)
$$x < -2 - \frac{\log{\left(4 \right)}}{\log{\left(2 \right)}}$$
x < -2 - log(4)/log(2)
Rapid solution 2
[src]
log(4)
(-oo, -2 - ------)
log(2)
$$x\ in\ \left(-\infty, -2 - \frac{\log{\left(4 \right)}}{\log{\left(2 \right)}}\right)$$
x in Interval.open(-oo, -2 - log(4)/log(2))