log((x+1),(x-2))<=0 inequation

Given the inequality:
$$\log{\left(x + 1 \right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(x + 1 \right)} = 0$$
Solve:
$$x_{1} = 0$$
$$x_{1} = 0$$
This roots
$$x_{1} = 0$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$\log{\left(x + 1 \right)} \leq 0$$
$$\log{\left(- \frac{1}{10} + 1 \right)} \leq 0$$

  log(9/10)        
--------------     
          /21\ <= 0
pi*I + log|--|     
          \10/     

Then
$$x \leq 0$$
no execute
the solution of our inequality is:
$$x \geq 0$$

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       x1