(3/10)^x>(9/100) inequation

Given the inequality:
$$\left(\frac{3}{10}\right)^{x} > \frac{9}{100}$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{3}{10}\right)^{x} = \frac{9}{100}$$
Solve:
Given the equation:
$$\left(\frac{3}{10}\right)^{x} = \frac{9}{100}$$
or
$$\left(\frac{3}{10}\right)^{x} - \frac{9}{100} = 0$$
or
$$\left(\frac{3}{10}\right)^{x} = \frac{9}{100}$$
or
$$\left(\frac{3}{10}\right)^{x} = \frac{9}{100}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{3}{10}\right)^{x}$$
we get
$$v - \frac{9}{100} = 0$$
or
$$v - \frac{9}{100} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{9}{100}$$
do backward replacement
$$\left(\frac{3}{10}\right)^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(\frac{3}{10} \right)}}$$
$$x_{1} = \frac{9}{100}$$
$$x_{1} = \frac{9}{100}$$
This roots
$$x_{1} = \frac{9}{100}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{9}{100}$$
=
$$- \frac{1}{100}$$
substitute to the expression
$$\left(\frac{3}{10}\right)^{x} > \frac{9}{100}$$
$$\frac{1}{\sqrt[100]{\frac{3}{10}}} > \frac{9}{100}$$

  99                
 ---                
 100 100____        
3   * \/ 10  > 9/100
------------        
     3              
        

the solution of our inequality is:
$$x < \frac{9}{100}$$

 _____          
      \    
-------ο-------
       x1