16^x-9<1 inequation

Given the inequality:
$$16^{x} - 9 < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$16^{x} - 9 = 1$$
Solve:
Given the equation:
$$16^{x} - 9 = 1$$
or
$$\left(16^{x} - 9\right) - 1 = 0$$
or
$$16^{x} = 10$$
or
$$16^{x} = 10$$
- this is the simplest exponential equation
Do replacement
$$v = 16^{x}$$
we get
$$v - 10 = 0$$
or
$$v - 10 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = 10$$
do backward replacement
$$16^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(16 \right)}}$$
$$x_{1} = 10$$
$$x_{1} = 10$$
This roots
$$x_{1} = 10$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 10$$
=
$$\frac{99}{10}$$
substitute to the expression
$$16^{x} - 9 < 1$$
$$-9 + 16^{\frac{99}{10}} < 1$$

                   3/5    
-9 + 549755813888*2    < 1
    

but

                   3/5    
-9 + 549755813888*2    > 1
    

Then
$$x < 10$$
no execute
the solution of our inequality is:
$$x > 10$$

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