(2^-x)<1/4 inequation

Given the inequality:
$$2^{- x} < \frac{1}{4}$$
To solve this inequality, we must first solve the corresponding equation:
$$2^{- x} = \frac{1}{4}$$
Solve:
Given the equation:
$$2^{- x} = \frac{1}{4}$$
or
$$- \frac{1}{4} + 2^{- x} = 0$$
or
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{4}$$
or
$$\left(\frac{1}{2}\right)^{x} = \frac{1}{4}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{2}\right)^{x}$$
we get
$$v - \frac{1}{4} = 0$$
or
$$v - \frac{1}{4} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{4}$$
do backward replacement
$$\left(\frac{1}{2}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = \frac{1}{4}$$
$$x_{1} = \frac{1}{4}$$
This roots
$$x_{1} = \frac{1}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{4}$$
=
$$\frac{3}{20}$$
substitute to the expression
$$2^{- x} < \frac{1}{4}$$
$$2^{- \frac{3}{20}} < \frac{1}{4}$$

 17      
 --      
 20      
2   < 1/4
---      
 2       
      

but

 17      
 --      
 20      
2   > 1/4
---      
 2       
      

Then
$$x < \frac{1}{4}$$
no execute
the solution of our inequality is:
$$x > \frac{1}{4}$$

         _____  
        /
-------ο-------
       x1