(3/8)^(2x+1)=<1 inequation

Given the inequality:
$$\left(\frac{3}{8}\right)^{2 x + 1} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\frac{3}{8}\right)^{2 x + 1} = 1$$
Solve:
Given the equation:
$$\left(\frac{3}{8}\right)^{2 x + 1} = 1$$
or
$$\left(\frac{3}{8}\right)^{2 x + 1} - 1 = 0$$
or
$$\frac{3 \left(\frac{9}{64}\right)^{x}}{8} = 1$$
or
$$\left(\frac{9}{64}\right)^{x} = \frac{8}{3}$$
- this is the simplest exponential equation
Do replacement
$$v = \left(\frac{9}{64}\right)^{x}$$
we get
$$v - \frac{8}{3} = 0$$
or
$$v - \frac{8}{3} = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = \frac{8}{3}$$
do backward replacement
$$\left(\frac{9}{64}\right)^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(\frac{9}{64} \right)}}$$
$$x_{1} = \frac{8}{3}$$
$$x_{1} = \frac{8}{3}$$
This roots
$$x_{1} = \frac{8}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{8}{3}$$
=
$$\frac{77}{30}$$
substitute to the expression
$$\left(\frac{3}{8}\right)^{2 x + 1} \leq 1$$
$$\left(\frac{3}{8}\right)^{1 + \frac{2 \cdot 77}{30}} \leq 1$$

     3/5  2/15     
729*2   *3         
-------------- <= 1
    524288         
     

the solution of our inequality is:
$$x \leq \frac{8}{3}$$

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       x1