(11/7)^x-4<1 inequation

Given the inequality:
$$\left(\frac{11}{7}\right)^{x} - 4 < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{11}{7}\right)^{x} - 4 = 1$$
Solve:
Given the equation:
$$\left(\frac{11}{7}\right)^{x} - 4 = 1$$
or
$$\left(\left(\frac{11}{7}\right)^{x} - 4\right) - 1 = 0$$
or
$$\left(\frac{11}{7}\right)^{x} = 5$$
or
$$\left(\frac{11}{7}\right)^{x} = 5$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{11}{7}\right)^{x}$$
we get
$$v - 5 = 0$$
or
$$v - 5 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 5$$
do backward replacement
$$\left(\frac{11}{7}\right)^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(\frac{11}{7} \right)}}$$
$$x_{1} = 5$$
$$x_{1} = 5$$
This roots
$$x_{1} = 5$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 5$$
=
$$\frac{49}{10}$$
substitute to the expression
$$\left(\frac{11}{7}\right)^{x} - 4 < 1$$
$$-4 + \left(\frac{11}{7}\right)^{\frac{49}{10}} < 1$$

           10___   9/10    
     14641*\/ 7 *11        
-4 + ------------------ < 1
           16807           
    

but

           10___   9/10    
     14641*\/ 7 *11        
-4 + ------------------ > 1
           16807           
    

Then
$$x < 5$$
no execute
the solution of our inequality is:
$$x > 5$$

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       x1