5^(1-x)>0,2 inequation

Given the inequality:
$$5^{1 - x} > \frac{1}{5}$$
To solve this inequality, we must first solve the corresponding equation:
$$5^{1 - x} = \frac{1}{5}$$
Solve:
Given the equation:
$$5^{1 - x} = \frac{1}{5}$$
or
$$5^{1 - x} - \frac{1}{5} = 0$$
or
$$5 \cdot 5^{- x} = \frac{1}{5}$$
or
$$\left(\frac{1}{5}\right)^{x} = \frac{1}{25}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{1}{5}\right)^{x}$$
we get
$$v - \frac{1}{25} = 0$$
or
$$v - \frac{1}{25} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{1}{25}$$
do backward replacement
$$\left(\frac{1}{5}\right)^{x} = v$$
or
$$x = - \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
$$x_{1} = \frac{1}{25}$$
$$x_{1} = \frac{1}{25}$$
This roots
$$x_{1} = \frac{1}{25}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{25}$$
=
$$- \frac{3}{50}$$
substitute to the expression
$$5^{1 - x} > \frac{1}{5}$$
$$5^{1 - - \frac{3}{50}} > \frac{1}{5}$$

   3/50      
5*5     > 1/5
      

the solution of our inequality is:
$$x < \frac{1}{25}$$

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