lg(2x+1)>lg5 inequation

Given the inequality:
$$\log{\left(2 x + 1 \right)} > \log{\left(5 \right)}$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(2 x + 1 \right)} = \log{\left(5 \right)}$$
Solve:
Given the equation
$$\log{\left(2 x + 1 \right)} = \log{\left(5 \right)}$$
$$\log{\left(2 x + 1 \right)} = \log{\left(5 \right)}$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$2 x + 1 = e^{\frac{\log{\left(5 \right)}}{1}}$$
simplify
$$2 x + 1 = 5$$
$$2 x = 4$$
$$x = 2$$
$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\log{\left(2 x + 1 \right)} > \log{\left(5 \right)}$$
$$\log{\left(1 + \frac{2 \cdot 19}{10} \right)} > \log{\left(5 \right)}$$

log(24/5) > log(5)

Then
$$x < 2$$
no execute
the solution of our inequality is:
$$x > 2$$

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