log2/5(2,5-2,5x)>-1 inequation

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

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

Expand expressions:

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

Reducing, you get:

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

Expand brackets in the left part

1 + log2/2 - x*log2/2 = 0

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

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

We get the answer: x = 1 + 2/log(2)
$$x_{1} = 1 + \frac{2}{\log{\left(2 \right)}}$$
$$x_{1} = 1 + \frac{2}{\log{\left(2 \right)}}$$
This roots
$$x_{1} = 1 + \frac{2}{\log{\left(2 \right)}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(1 + \frac{2}{\log{\left(2 \right)}}\right)$$
=
$$\frac{9}{10} + \frac{2}{\log{\left(2 \right)}}$$
substitute to the expression
$$\frac{\log{\left(2 \right)}}{5} \left(\frac{5}{2} - \frac{5 x}{2}\right) > -1$$
$$\frac{\log{\left(2 \right)}}{5} \left(\frac{5}{2} - \frac{5 \left(\frac{9}{10} + \frac{2}{\log{\left(2 \right)}}\right)}{2}\right) > -1$$

/1     5   \            
|- - ------|*log(2)     
\4   log(2)/        > -1
-------------------     
         5              

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

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       x1