log0,2x>1 inequation

Given the inequality:
$$\log{\left(0.2 x \right)} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(0.2 x \right)} = 1$$
Solve:
Given the equation
$$\log{\left(0.2 x \right)} = 1$$
$$\log{\left(0.2 x \right)} = 1$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$0.2 x = e^{1^{-1}}$$
simplify
$$0.2 x = e$$
$$x = 5 e$$
$$x_{1} = 13.5914091422952$$
$$x_{1} = 13.5914091422952$$
This roots
$$x_{1} = 13.5914091422952$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 13.5914091422952$$
=
$$13.4914091422952$$
substitute to the expression
$$\log{\left(0.2 x \right)} > 1$$
$$\log{\left(0.2 \cdot 13.4914091422952 \right)} > 1$$

0.992615210617444 > 1

Then
$$x < 13.5914091422952$$
no execute
the solution of our inequality is:
$$x > 13.5914091422952$$

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