log5(3+8*x)>0 inequation

Given the inequality:
$$\frac{\log{\left(8 x + 3 \right)}}{\log{\left(5 \right)}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(8 x + 3 \right)}}{\log{\left(5 \right)}} = 0$$
Solve:
Given the equation
$$\frac{\log{\left(8 x + 3 \right)}}{\log{\left(5 \right)}} = 0$$
$$\frac{\log{\left(8 x + 3 \right)}}{\log{\left(5 \right)}} = 0$$
Let's divide both parts of the equation by the multiplier of log =1/log(5)
$$\log{\left(8 x + 3 \right)} = 0$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$8 x + 3 = e^{\frac{0}{\frac{1}{\log{\left(5 \right)}}}}$$
simplify
$$8 x + 3 = 1$$
$$8 x = -2$$
$$x = - \frac{1}{4}$$
$$x_{1} = - \frac{1}{4}$$
$$x_{1} = - \frac{1}{4}$$
This roots
$$x_{1} = - \frac{1}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{4} + - \frac{1}{10}$$
=
$$- \frac{7}{20}$$
substitute to the expression
$$\frac{\log{\left(8 x + 3 \right)}}{\log{\left(5 \right)}} > 0$$
$$\frac{\log{\left(\frac{\left(-7\right) 8}{20} + 3 \right)}}{\log{\left(5 \right)}} > 0$$

-1 > 0

Then
$$x < - \frac{1}{4}$$
no execute
the solution of our inequality is:
$$x > - \frac{1}{4}$$

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