3^(x+1)+3>0 inequation

Given the inequality:
$$3^{x + 1} + 3 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$3^{x + 1} + 3 = 0$$
Solve:
Given the equation:
$$3^{x + 1} + 3 = 0$$
or
$$3^{x + 1} + 3 = 0$$
or
$$3 \cdot 3^{x} = -3$$
or
$$3^{x} = -1$$
- this is the simplest exponential equation
Do replacement
$$v = 3^{x}$$
we get
$$v + 1 = 0$$
or
$$v + 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = -1$$
do backward replacement
$$3^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(3 \right)}}$$
$$x_{1} = -1$$
$$x_{1} = -1$$
This roots
$$x_{1} = -1$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$3^{x + 1} + 3 > 0$$
$$3^{- \frac{11}{10} + 1} + 3 > 0$$

     9/10    
    3        
3 + ----- > 0
      3      
    

the solution of our inequality is:
$$x < -1$$

 _____          
      \    
-------ο-------
       x1