0,2^(x+1)<=1/25 inequation

Given the inequality:
$$\left(\frac{1}{5}\right)^{x + 1} \leq \frac{1}{25}$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{5}\right)^{x + 1} = \frac{1}{25}$$
Solve:
Given the equation:
$$\left(\frac{1}{5}\right)^{x + 1} = \frac{1}{25}$$
or
$$\left(\frac{1}{5}\right)^{x + 1} - \frac{1}{25} = 0$$
or
$$\frac{5^{- x}}{5} = \frac{1}{25}$$
or
$$\left(\frac{1}{5}\right)^{x} = \frac{1}{5}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{5}\right)^{x}$$
we get
$$v - \frac{1}{5} = 0$$
or
$$v - \frac{1}{5} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{5}$$
do backward replacement
$$\left(\frac{1}{5}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
$$x_{1} = \frac{1}{5}$$
$$x_{1} = \frac{1}{5}$$
This roots
$$x_{1} = \frac{1}{5}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{5}$$
=
$$\frac{1}{10}$$
substitute to the expression
$$\left(\frac{1}{5}\right)^{x + 1} \leq \frac{1}{25}$$
$$\left(\frac{1}{5}\right)^{\frac{1}{10} + 1} \leq \frac{1}{25}$$

 9/10        
5            
----- <= 1/25
  25         
        

but

 9/10        
5            
----- >= 1/25
  25         
        

Then
$$x \leq \frac{1}{5}$$
no execute
the solution of our inequality is:
$$x \geq \frac{1}{5}$$

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