Mister Exam
2^(2*x)-7*(2^x)-8>0 inequation
The solution
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2*x x 2 - 7*2 - 8 > 0
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 > 0$$
2^(2*x) - 7*2^x - 8 > 0
Detail solution
Given the inequality:
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
Solve:
Given the equation:
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
or
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 7 v - 8 = 0$$
or
$$v^{2} - 7 v - 8 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = -8$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 8$$
$$v_{2} = -1$$
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = 8$$
$$x_{2} = -1$$
$$x_{1} = 8$$
$$x_{2} = -1$$
This roots
$$x_{2} = -1$$
$$x_{1} = 8$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 > 0$$
$$-8 + \left(- \frac{7}{2^{\frac{11}{10}}} + 2^{\frac{\left(-11\right) 2}{10}}\right) > 0$$
Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < 8$$
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
Solve:
Given the equation:
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
or
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 7 v - 8 = 0$$
or
$$v^{2} - 7 v - 8 = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = -8$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (1) * (-8) = 81
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 8$$
$$v_{2} = -1$$
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
$$x_{1} = 8$$
$$x_{2} = -1$$
$$x_{1} = 8$$
$$x_{2} = -1$$
This roots
$$x_{2} = -1$$
$$x_{1} = 8$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\left(2^{2 x} - 7 \cdot 2^{x}\right) - 8 > 0$$
$$-8 + \left(- \frac{7}{2^{\frac{11}{10}}} + 2^{\frac{\left(-11\right) 2}{10}}\right) > 0$$
9/10 4/5
7*2 2
-8 - ------- + ---- > 0
4 8
Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < 8$$
_____
/ \
-------ο-------ο-------
x2 x1
Solving inequality on a graph
Rapid solution
[src]
/ log(8) \ And|x < oo, ------ < x| \ log(2) /
$$x < \infty \wedge \frac{\log{\left(8 \right)}}{\log{\left(2 \right)}} < x$$
(x < oo)∧(log(8)/log(2) < x)
Rapid solution 2
[src]
log(8) (------, oo) log(2)
$$x\ in\ \left(\frac{\log{\left(8 \right)}}{\log{\left(2 \right)}}, \infty\right)$$
x in Interval.open(log(8)/log(2), oo)