log(5)(2x−1)<1 inequation

Given the inequality:
$$\left(2 x - 1\right) \log{\left(5 \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2 x - 1\right) \log{\left(5 \right)} = 1$$
Solve:
Given the equation:

log(5)*(2*x-1) = 1

Expand expressions:

-log(5) + 2*x*log(5) = 1

Reducing, you get:

-1 - log(5) + 2*x*log(5) = 0

Expand brackets in the left part

-1 - log5 + 2*x*log5 = 0

Move free summands (without x)
from left part to right part, we given:
$$2 x \log{\left(5 \right)} - \log{\left(5 \right)} = 1$$
Divide both parts of the equation by (-log(5) + 2*x*log(5))/x

x = 1 / ((-log(5) + 2*x*log(5))/x)

We get the answer: x = (1 + log(5))/(2*log(5))
$$x_{1} = \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$
$$x_{1} = \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$
This roots
$$x_{1} = \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$
=
$$- \frac{1}{10} + \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$
substitute to the expression
$$\left(2 x - 1\right) \log{\left(5 \right)} < 1$$
$$\left(-1 + 2 \left(- \frac{1}{10} + \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}\right)\right) \log{\left(5 \right)} < 1$$

/  6   1 + log(5)\           
|- - + ----------|*log(5) < 1
\  5     log(5)  /           

the solution of our inequality is:
$$x < \frac{1 + \log{\left(5 \right)}}{2 \log{\left(5 \right)}}$$

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