Given the inequality:
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} < \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
Solve:
Given the equation
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(2 x + 1 \right)} = \log{\left(5 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$2 x + 1 = e^{\frac{\frac{1}{\log{\left(3 \right)}} \log{\left(5 \right)}}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$2 x + 1 = 5$$
$$2 x = 4$$
$$x = 2$$
$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\frac{\log{\left(2 x + 1 \right)}}{\log{\left(3 \right)}} < \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
$$\frac{\log{\left(1 + \frac{2 \cdot 19}{10} \right)}}{\log{\left(3 \right)}} < \frac{\log{\left(5 \right)}}{\log{\left(3 \right)}}$$
log(24/5) log(5)
--------- < ------
log(3) log(3)
the solution of our inequality is:
$$x < 2$$
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