Mister Exam
sqrt(2^x)<=128 inequation
The solution
Detail solution
Given the inequality:
$$\sqrt{2^{x}} \leq 128$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2^{x}} = 128$$
Solve:
$$x_{1} = 14$$
$$x_{1} = 14$$
This roots
$$x_{1} = 14$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 14$$
=
$$\frac{139}{10}$$
substitute to the expression
$$\sqrt{2^{x}} \leq 128$$
$$\sqrt{2^{\frac{139}{10}}} \leq 128$$
the solution of our inequality is:
$$x \leq 14$$
$$\sqrt{2^{x}} \leq 128$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2^{x}} = 128$$
Solve:
$$x_{1} = 14$$
$$x_{1} = 14$$
This roots
$$x_{1} = 14$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 14$$
=
$$\frac{139}{10}$$
substitute to the expression
$$\sqrt{2^{x}} \leq 128$$
$$\sqrt{2^{\frac{139}{10}}} \leq 128$$
___ 9/20
64*\/ 2 *2 <= 128
the solution of our inequality is:
$$x \leq 14$$
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