log3x>9 inequation

Given the inequality:
$$\log{\left(3 x \right)} > 9$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(3 x \right)} = 9$$
Solve:
Given the equation
$$\log{\left(3 x \right)} = 9$$
$$\log{\left(3 x \right)} = 9$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$3 x + 0 = e^{\frac{9}{1}}$$
simplify
$$3 x = e^{9}$$
$$x = \frac{e^{9}}{3}$$
$$x_{1} = \frac{e^{9}}{3}$$
$$x_{1} = \frac{e^{9}}{3}$$
This roots
$$x_{1} = \frac{e^{9}}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{e^{9}}{3}$$
=
$$- \frac{1}{10} + \frac{e^{9}}{3}$$
substitute to the expression
$$\log{\left(3 x \right)} > 9$$
$$\log{\left(3 \left(- \frac{1}{10} + \frac{e^{9}}{3}\right) \right)} > 9$$

   /  3     9\    
log|- -- + e | > 9
   \  10     /    

Then
$$x < \frac{e^{9}}{3}$$
no execute
the solution of our inequality is:
$$x > \frac{e^{9}}{3}$$

         _____  
        /
-------ο-------
       x_1