loq⅓(x-1)>-2 inequation

Given the inequality:
$$\left(x - 1\right) \log{\left(\frac{1}{3} \right)} > -2$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x - 1\right) \log{\left(\frac{1}{3} \right)} = -2$$
Solve:
Given the equation:

log(1/3)*(x-1) = -2

Expand expressions:

-x*log(3) + log(3) = -2

Reducing, you get:

2 - x*log(3) + log(3) = 0

Expand brackets in the left part

2 - x*log3 + log3 = 0

Move free summands (without x)
from left part to right part, we given:
$$- x \log{\left(3 \right)} + \log{\left(3 \right)} = -2$$
Divide both parts of the equation by (-x*log(3) + log(3))/x

x = -2 / ((-x*log(3) + log(3))/x)

We get the answer: x = 1 + 2/log(3)
$$x_{1} = 1 + \frac{2}{\log{\left(3 \right)}}$$
$$x_{1} = 1 + \frac{2}{\log{\left(3 \right)}}$$
This roots
$$x_{1} = 1 + \frac{2}{\log{\left(3 \right)}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(1 + \frac{2}{\log{\left(3 \right)}}\right)$$
=
$$\frac{9}{10} + \frac{2}{\log{\left(3 \right)}}$$
substitute to the expression
$$\left(x - 1\right) \log{\left(\frac{1}{3} \right)} > -2$$
$$\left(-1 + \left(\frac{9}{10} + \frac{2}{\log{\left(3 \right)}}\right)\right) \log{\left(\frac{1}{3} \right)} > -2$$

 /  1      2   \            
-|- -- + ------|*log(3) > -2
 \  10   log(3)/            

the solution of our inequality is:
$$x < 1 + \frac{2}{\log{\left(3 \right)}}$$

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