log3(2*x+1)<3 inequation

Given the inequality:
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} < 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = 3$$
Solve:
Given the equation
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = 3$$
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = 3$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(2 x + 1 \right)} = 3 \log{\left(3 \right)}$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$2 x + 1 = e^{\frac{3}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$2 x + 1 = 27$$
$$2 x = 26$$
$$x = 13$$
$$x_{1} = 13$$
$$x_{1} = 13$$
This roots
$$x_{1} = 13$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 13$$
=
$$\frac{129}{10}$$
substitute to the expression
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} < 3$$
$$\frac{\log{\left(1 + \frac{2 \cdot 129}{10} \right)}}{\log{\left(3 \right)}} < 3$$

log(134/5)    
---------- < 3
  log(3)      

the solution of our inequality is:
$$x < 13$$

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