log(3-2x)<2log5-1 inequation

Given the inequality:
$$\log{\left(3 - 2 x \right)} < -1 + 2 \log{\left(5 \right)}$$
To solve this inequality, we must first solve the corresponding equation:
$$\log{\left(3 - 2 x \right)} = -1 + 2 \log{\left(5 \right)}$$
Solve:
Given the equation
$$\log{\left(3 - 2 x \right)} = -1 + 2 \log{\left(5 \right)}$$
$$\log{\left(3 - 2 x \right)} = -1 + 2 \log{\left(5 \right)}$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$3 - 2 x = e^{\frac{-1 + 2 \log{\left(5 \right)}}{1}}$$
simplify
$$3 - 2 x = \frac{25}{e}$$
$$- 2 x = -3 + \frac{25}{e}$$
$$x = \frac{3}{2} - \frac{25}{2 e}$$
$$x_{1} = \frac{-25 + 3 e}{2 e}$$
$$x_{1} = \frac{-25 + 3 e}{2 e}$$
This roots
$$x_{1} = \frac{-25 + 3 e}{2 e}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\frac{-25 + 3 e}{2 e^{1}} + - \frac{1}{10}$$
=
$$\frac{-25 + 3 e}{2 e} - \frac{1}{10}$$
substitute to the expression
$$\log{\left(3 - 2 x \right)} < -1 + 2 \log{\left(5 \right)}$$
$$\log{\left(3 - 2 \left(\frac{-25 + 3 e}{2 e^{1}} - \frac{1}{10}\right) \right)} < -1 + 2 \log{\left(5 \right)}$$

   /16                -1\                
log|-- - (-25 + 3*E)*e  | < -1 + 2*log(5)
   \5                   /                

but

   /16                -1\                
log|-- - (-25 + 3*E)*e  | > -1 + 2*log(5)
   \5                   /                

Then
$$x < \frac{-25 + 3 e}{2 e}$$
no execute
the solution of our inequality is:
$$x > \frac{-25 + 3 e}{2 e}$$

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