3^(x+2)>81 inequation

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:
$$3^{x + 2} = 81$$
or
$$3^{x + 2} - 81 = 0$$
or
$$9 \cdot 3^{x} = 81$$
or
$$3^{x} = 9$$
- this is the simplest exponential equation
Do replacement
$$v = 3^{x}$$
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
$$3^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(3 \right)}}$$
$$x_{1} = 9$$
$$x_{1} = 9$$
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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