6^(x^2-9)>1 inequation

Given the inequality:
$$6^{x^{2} - 9} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$6^{x^{2} - 9} = 1$$
Solve:
$$x_{1} = -3$$
$$x_{2} = 3$$
$$x_{1} = -3$$
$$x_{2} = 3$$
This roots
$$x_{1} = -3$$
$$x_{2} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$6^{x^{2} - 9} > 1$$
$$6^{-9 + \left(- \frac{31}{10}\right)^{2}} > 1$$

  61    
 ---    
 100 > 1
6       
    

one of the solutions of our inequality is:
$$x < -3$$

 _____           _____          
      \         /
-------ο-------ο-------
       x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -3$$
$$x > 3$$