1/4^(x+4)>8√2 inequation

Given the inequality:
$$\left(\frac{1}{4}\right)^{x + 4} > 8 \sqrt{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{4}\right)^{x + 4} = 8 \sqrt{2}$$
Solve:
Given the equation:
$$\left(\frac{1}{4}\right)^{x + 4} = 8 \sqrt{2}$$
or
$$\left(\frac{1}{4}\right)^{x + 4} - 8 \sqrt{2} = 0$$
or
$$\frac{4^{- x}}{256} = 8 \sqrt{2}$$
or
$$\left(\frac{1}{4}\right)^{x} = 2048 \sqrt{2}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{4}\right)^{x}$$
we get
$$v - 2048 \sqrt{2} = 0$$
or
$$v - 2048 \sqrt{2} = 0$$
Expand brackets in the left part

v - 2048*sqrt2 = 0

Divide both parts of the equation by (v - 2048*sqrt(2))/v

v = 0 / ((v - 2048*sqrt(2))/v)

do backward replacement
$$\left(\frac{1}{4}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(4 \right)}}$$
$$x_{1} = 2048 \sqrt{2}$$
$$x_{1} = 2048 \sqrt{2}$$
This roots
$$x_{1} = 2048 \sqrt{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2048 \sqrt{2}$$
=
$$- \frac{1}{10} + 2048 \sqrt{2}$$
substitute to the expression
$$\left(\frac{1}{4}\right)^{x + 4} > 8 \sqrt{2}$$
$$\left(\frac{1}{4}\right)^{4 + \left(- \frac{1}{10} + 2048 \sqrt{2}\right)} > 8 \sqrt{2}$$

   39          ___          
 - -- - 2048*\/ 2        ___
   10              > 8*\/ 2 
4                    
          

Then
$$x < 2048 \sqrt{2}$$
no execute
the solution of our inequality is:
$$x > 2048 \sqrt{2}$$

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