log_x-1((x+1)/5)<=0 inequation

Given the inequality:
$$- \frac{1 \left(x + 1\right)}{5} + \log{\left(_x \right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- \frac{1 \left(x + 1\right)}{5} + \log{\left(_x \right)} = 0$$
Solve:
Given the equation:
$$- \frac{1 \left(x + 1\right)}{5} + \log{\left(_x \right)} = 0$$
transform:
$$x - 5 \log{\left(_x \right)} + 1 = 0$$
Expand brackets in the left part

1 + x - 5*logx = 0

Move free summands (without x)
from left part to right part, we given:
$$x - 5 \log{\left(_x \right)} = -1$$
Divide both parts of the equation by (x - 5*log(_x))/x

x = -1 / ((x - 5*log(_x))/x)

$$x_{1} = 5 \log{\left(_x \right)} - 1$$
$$x_{1} = 5 \log{\left(_x \right)} - 1$$
This roots
$$x_{1} = 5 \log{\left(_x \right)} - 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(5 \log{\left(_x \right)} - 1\right) - \frac{1}{10}$$
=
$$5 \log{\left(_x \right)} - \frac{11}{10}$$
substitute to the expression
$$- \frac{1 \left(x + 1\right)}{5} + \log{\left(_x \right)} \leq 0$$
$$- \frac{1 \left(\left(5 \log{\left(_x \right)} - \frac{11}{10}\right) + 1\right)}{5} + \log{\left(_x \right)} \leq 0$$

1/50 <= 0

but

1/50 >= 0

Then
$$x \leq 5 \log{\left(_x \right)} - 1$$
no execute
the solution of our inequality is:
$$x \geq 5 \log{\left(_x \right)} - 1$$

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