(1/5)^(x-1)-(1/5)^x<=100 inequation

Given the inequality:
$$\left(\frac{1}{5}\right)^{x - 1} - \left(\frac{1}{5}\right)^{x} \leq 100$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{1}{5}\right)^{x - 1} - \left(\frac{1}{5}\right)^{x} = 100$$
Solve:
Given the equation:
$$\left(\frac{1}{5}\right)^{x - 1} - \left(\frac{1}{5}\right)^{x} = 100$$
or
$$\left(\left(\frac{1}{5}\right)^{x - 1} - \left(\frac{1}{5}\right)^{x}\right) - 100 = 0$$
Do replacement
$$v = \left(\frac{1}{5}\right)^{x}$$
we get
$$4 v - 100 = 0$$
or
$$4 v - 100 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$4 v = 100$$
Divide both parts of the equation by 4

v = 100 / (4)

do backward replacement
$$\left(\frac{1}{5}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
$$x_{1} = 25$$
$$x_{1} = 25$$
This roots
$$x_{1} = 25$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 25$$
=
$$\frac{249}{10}$$
substitute to the expression
$$\left(\frac{1}{5}\right)^{x - 1} - \left(\frac{1}{5}\right)^{x} \leq 100$$
$$- \left(\frac{1}{5}\right)^{\frac{249}{10}} + \left(\frac{1}{5}\right)^{-1 + \frac{249}{10}} \leq 100$$

       10___             
     4*\/ 5              
------------------ <= 100
298023223876953125       
       

the solution of our inequality is:
$$x \leq 25$$

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