log31(31x+2)<=0 inequation

Given the inequality:
$$\frac{\log{\left(31 x + 2 \right)}}{\log{\left(31 \right)}} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(31 x + 2 \right)}}{\log{\left(31 \right)}} = 0$$
Solve:
Given the equation
$$\frac{\log{\left(31 x + 2 \right)}}{\log{\left(31 \right)}} = 0$$
$$\frac{\log{\left(31 x + 2 \right)}}{\log{\left(31 \right)}} = 0$$
Let's divide both parts of the equation by the multiplier of log =1/log(31)
$$\log{\left(31 x + 2 \right)} = 0$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$31 x + 2 = e^{\frac{0}{\frac{1}{\log{\left(31 \right)}}}}$$
simplify
$$31 x + 2 = 1$$
$$31 x = -1$$
$$x = - \frac{1}{31}$$
$$x_{1} = - \frac{1}{31}$$
$$x_{1} = - \frac{1}{31}$$
This roots
$$x_{1} = - \frac{1}{31}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} - \frac{1}{31}$$
=
$$- \frac{41}{310}$$
substitute to the expression
$$\frac{\log{\left(31 x + 2 \right)}}{\log{\left(31 \right)}} \leq 0$$
$$\frac{\log{\left(31 \left(- \frac{41}{310}\right) + 2 \right)}}{\log{\left(31 \right)}} \leq 0$$

          /21\     
pi*I + log|--|     
          \10/ <= 0
--------------     
   log(31)         

Then
$$x \leq - \frac{1}{31}$$
no execute
the solution of our inequality is:
$$x \geq - \frac{1}{31}$$

         _____  
        /
-------•-------
       x_1