log5(3x-1)>0 inequation

Given the inequality:
$$\frac{\log{\left(3 x - 1 \right)}}{\log{\left(5 \right)}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(3 x - 1 \right)}}{\log{\left(5 \right)}} = 0$$
Solve:
Given the equation
$$\frac{\log{\left(3 x - 1 \right)}}{\log{\left(5 \right)}} = 0$$
$$\frac{\log{\left(3 x - 1 \right)}}{\log{\left(5 \right)}} = 0$$
Let's divide both parts of the equation by the multiplier of log =1/log(5)
$$\log{\left(3 x - 1 \right)} = 0$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$3 x - 1 = e^{\frac{0}{\frac{1}{\log{\left(5 \right)}}}}$$
simplify
$$3 x - 1 = 1$$
$$3 x = 2$$
$$x = \frac{2}{3}$$
$$x_{1} = \frac{2}{3}$$
$$x_{1} = \frac{2}{3}$$
This roots
$$x_{1} = \frac{2}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{2}{3}$$
=
$$\frac{17}{30}$$
substitute to the expression
$$\frac{\log{\left(3 x - 1 \right)}}{\log{\left(5 \right)}} > 0$$
$$\frac{\log{\left(-1 + \frac{3 \cdot 17}{30} \right)}}{\log{\left(5 \right)}} > 0$$

log(7/10)    
--------- > 0
  log(5)     

Then
$$x < \frac{2}{3}$$
no execute
the solution of our inequality is:
$$x > \frac{2}{3}$$

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