0,5^x>0,5 inequation

Given the inequality:
$$\left(\frac{1}{2}\right)^{x} > \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{2}$$
Solve:
Given the equation:
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{2}$$
or
$$- \frac{1}{2} + \left(\frac{1}{2}\right)^{x} = 0$$
or
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{2}$$
or
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{2}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{2}\right)^{x}$$
we get
$$v - \frac{1}{2} = 0$$
or
$$v - \frac{1}{2} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{2}$$
do backward replacement
$$\left(\frac{1}{2}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = \frac{1}{2}$$
$$x_{1} = \frac{1}{2}$$
This roots
$$x_{1} = \frac{1}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{2}$$
=
$$\frac{2}{5}$$
substitute to the expression
$$\left(\frac{1}{2}\right)^{x} > \frac{1}{2}$$
$$\left(\frac{1}{2}\right)^{\frac{2}{5}} > \frac{1}{2}$$

 3/5      
2         
---- > 1/2
 2        
      

the solution of our inequality is:
$$x < \frac{1}{2}$$

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