log(7)(15-x)<1 inequation

Given the inequality:
$$\left(15 - x\right) \log{\left(7 \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(15 - x\right) \log{\left(7 \right)} = 1$$
Solve:
Given the equation:

log(7)*(15-x) = 1

Expand expressions:

15*log(7) - x*log(7) = 1

Reducing, you get:

-1 + 15*log(7) - x*log(7) = 0

Expand brackets in the left part

-1 + 15*log7 - x*log7 = 0

Move free summands (without x)
from left part to right part, we given:
$$- x \log{\left(7 \right)} + 15 \log{\left(7 \right)} = 1$$
Divide both parts of the equation by (15*log(7) - x*log(7))/x

x = 1 / ((15*log(7) - x*log(7))/x)

We get the answer: x = (-1 + log(4747561509943))/log(7)
$$x_{1} = \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
$$x_{1} = \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
This roots
$$x_{1} = \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
=
$$- \frac{1}{10} + \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
substitute to the expression
$$\left(15 - x\right) \log{\left(7 \right)} < 1$$
$$\left(15 - \left(- \frac{1}{10} + \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}\right)\right) \log{\left(7 \right)} < 1$$

/151   -1 + log(4747561509943)\           
|--- - -----------------------|*log(7) < 1
\ 10            log(7)        /           

but

/151   -1 + log(4747561509943)\           
|--- - -----------------------|*log(7) > 1
\ 10            log(7)        /           

Then
$$x < \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$
no execute
the solution of our inequality is:
$$x > \frac{-1 + \log{\left(4747561509943 \right)}}{\log{\left(7 \right)}}$$

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