Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation sin^2((П/4)-x)=sin^2((П/4)+x)
- Equation log2(56)-(1/2log2(49))=x
-
Equation 4^x-29*2^x+168=0
-
Equation 31+25x+2x^2=7x-9
- Express {x} in terms of y where:
- 5*x-5*y=20
- 8*x-7*y=-13
- 15*x-15*y=16
- 9*x+18*y=6
-
Identical expressions
- four ^x- twenty-nine * two ^x+ one hundred and sixty-eight = zero
- 4 to the power of x minus 29 multiply by 2 to the power of x plus 168 equally 0
- four to the power of x minus twenty minus nine multiply by two to the power of x plus one hundred and sixty minus eight equally zero
- 4x-29*2x+168=0
- 4^x-292^x+168=0
- 4x-292x+168=0
- 4^x-29*2^x+168=O
-
Similar expressions
- 4^x-29*2^x-168=0
- 4^x+29*2^x+168=0
4^x-29*2^x+168=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$- 29 \cdot 2^{x} + 4^{x} + 168 = 0$$
or
$$\left(- 29 \cdot 2^{x} + 4^{x} + 168\right) + 0 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 29 v + 168 = 0$$
or
$$v^{2} - 29 v + 168 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -29$$
$$c = 168$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 168 + \left(-29\right)^{2} = 169$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 21$$
Simplify
$$v_{2} = 8$$
Simplify
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(8 \right)}}{\log{\left(2 \right)}} = 3$$
$$x_{2} = \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}$$
$$- 29 \cdot 2^{x} + 4^{x} + 168 = 0$$
or
$$\left(- 29 \cdot 2^{x} + 4^{x} + 168\right) + 0 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 29 v + 168 = 0$$
or
$$v^{2} - 29 v + 168 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -29$$
$$c = 168$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 168 + \left(-29\right)^{2} = 169$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 21$$
Simplify
$$v_{2} = 8$$
Simplify
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(8 \right)}}{\log{\left(2 \right)}} = 3$$
$$x_{2} = \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}$$
Sum and product of roots
[src]
sum
log(21)
3 + -------
log(2)
$$\left(3\right) + \left(\frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}\right)$$
=
log(21)
3 + -------
log(2)
$$3 + \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}$$
product
log(21)
3 * -------
log(2)
$$\left(3\right) * \left(\frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}\right)$$
=
3*log(21) --------- log(2)
$$\frac{3 \log{\left(21 \right)}}{\log{\left(2 \right)}}$$
Rapid solution
[src]
x_1 = 3
$$x_{1} = 3$$
log(21)
x_2 = -------
log(2)
$$x_{2} = \frac{\log{\left(21 \right)}}{\log{\left(2 \right)}}$$
The graph