3^(2-4*x)<=9 inequation

Given the inequality:
$$3^{2 - 4 x} \leq 9$$
To solve this inequality, we must first solve the corresponding equation:
$$3^{2 - 4 x} = 9$$
Solve:
Given the equation:
$$3^{2 - 4 x} = 9$$
or
$$3^{2 - 4 x} - 9 = 0$$
or
$$9 \cdot 81^{- x} = 9$$
or
$$\left(\frac{1}{81}\right)^{x} = 1$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{81}\right)^{x}$$
we get
$$v - 1 = 0$$
or
$$v - 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 1$$
do backward replacement
$$\left(\frac{1}{81}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(81 \right)}}$$
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$3^{2 - 4 x} \leq 9$$
$$3^{2 - \frac{4 \cdot 9}{10}} \leq 9$$

 2/5     
3        
---- <= 9
 9       
     

the solution of our inequality is:
$$x \leq 1$$

 _____          
      \    
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       x1