Mister Exam

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How to use it?
Inequation:
(x+3)^2>(x−23)^2
3^(x+2)>81
(x^2+5*x)/(x-6)>0
3^x>7*2^x
Graphing y =:
3^(x+2)
Sum of series:
81
Identical expressions
three ^(x+ two)> eighty-one
3 to the power of (x plus 2) greater than 81
three to the power of (x plus two) greater than eighty minus one
3(x+2)>81
3x+2>81
3^x+2>81
Similar expressions
(3*9^x)-(8*15^x)+25^(x+0,5)>=0
3^(x-2)>81
(45*2^x-90+45*2^(-x))*1/(2^x+2+2^(-x))-(21*2^x+21)*1/(2^x+1)<=(2^x+3-8)*1/(2^x+1)

3^(x+2)>81 inequation

You have entered [src]

 x + 2     
3      > 81

3^{x + 2} > 81

3^(x + 2) > 81

Detail solution

Given the inequality:

3^{x + 2} > 81

To solve this inequality, we must first solve the corresponding equation:

3^{x + 2} = 81

Solve:
Given the equation:

3^{x + 2} = 81

3^{x + 2} - 81 = 0

9 \cdot 3^{x} = 81

3^{x} = 9

- this is the simplest exponential equation
Do replacement

v = 3^{x}

we get

v - 9 = 0

v - 9 = 0

Move free summands (without v)
from left part to right part, we given:

v = 9

do backward replacement

3^{x} = v

x = \frac{\log{\left(v \right)}}{\log{\left(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

3^{x + 2} > 81

3^{2 + \frac{89}{10}} > 81

       9/10     
59049*3     > 81

the solution of our inequality is:

x < 9

 _____          
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       x1

Solving inequality on a graph

Rapid solution 2 [src]

(2, oo)

x\ in\ \left(2, \infty\right)

x in Interval.open(2, oo)

Rapid solution [src]

2 < x

2 < x

2 < x