lgx>=1 inequation

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

log(v)=p

By definition log

v=e^p

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

log(-1/10 + E) >= 1

but

log(-1/10 + E) < 1

Then
$$x \leq e$$
no execute
the solution of our inequality is:
$$x \geq e$$

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