Given the inequality:
$$\frac{\left(27 - 3^{x}\right) \left(7^{x} - 7\right)}{5^{x}} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(27 - 3^{x}\right) \left(7^{x} - 7\right)}{5^{x}} = 0$$
Solve:
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{1} = 1$$
$$x_{2} = 3$$
This roots
$$x_{1} = 1$$
$$x_{2} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\frac{\left(27 - 3^{x}\right) \left(7^{x} - 7\right)}{5^{x}} \geq 0$$
$$\frac{\left(-7 + 7^{\frac{9}{10}}\right) \left(27 - 3^{\frac{9}{10}}\right)}{5^{\frac{9}{10}}} \geq 0$$
10___ / 9/10\ / 9/10\
\/ 5 *\-7 + 7 /*\27 - 3 /
------------------------------- >= 0
5
but
10___ / 9/10\ / 9/10\
\/ 5 *\-7 + 7 /*\27 - 3 /
------------------------------- < 0
5
Then
$$x \leq 1$$
no execute
one of the solutions of our inequality is:
$$x \geq 1 \wedge x \leq 3$$
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