lg(2x-7)>0 inequation

Given the inequality:
$$\log{\left(2 x - 7 \right)} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(2 x - 7 \right)} = 0$$
Solve:
Given the equation
$$\log{\left(2 x - 7 \right)} = 0$$
$$\log{\left(2 x - 7 \right)} = 0$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$2 x - 7 = e^{\frac{0}{1}}$$
simplify
$$2 x - 7 = 1$$
$$2 x = 8$$
$$x = 4$$
$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$\log{\left(2 x - 7 \right)} > 0$$
$$\log{\left(\left(-1\right) 7 + 2 \cdot \frac{39}{10} \right)} > 0$$

log(4/5) > 0

Then
$$x < 4$$
no execute
the solution of our inequality is:
$$x > 4$$

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