Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x+12=5
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Equation x^3=4
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Equation 0,144:(3,4-x)=2,4
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Equation x^4=(x-56)^2
- Express {x} in terms of y where:
- 15*x-5*y=-5
- 18*x+20*y=-2
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Identical expressions
- - three ^(x+ two)+ nine ^x+ fourteen = zero
- minus 3 to the power of (x plus 2) plus 9 to the power of x plus 14 equally 0
- minus three to the power of (x plus two) plus nine to the power of x plus fourteen equally zero
- -3(x+2)+9x+14=0
- -3x+2+9x+14=0
- -3^x+2+9^x+14=0
- -3^(x+2)+9^x+14=O
-
Similar expressions
- 3^(x+2)+9^x+14=0
- -3^(x+2)+9^x-14=0
- -3^(x-2)+9^x+14=0
- -3^(x+2)-9^x+14=0
-3^(x+2)+9^x+14=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$- 3^{x + 2} + 9^{x} + 14 = 0$$
or
$$\left(- 3^{x + 2} + 9^{x} + 14\right) + 0 = 0$$
Do replacement
$$v = 3^{x}$$
we get
$$v^{2} - 9 v + 14 = 0$$
or
$$v^{2} - 9 v + 14 = 0$$
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
$$a = 1$$
$$b = -9$$
$$c = 14$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 14 + \left(-9\right)^{2} = 25$$
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
$$v_{1} = 7$$
Simplify
$$v_{2} = 2$$
Simplify
do backward replacement
$$3^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(3 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}$$
$$x_{2} = \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}$$
$$- 3^{x + 2} + 9^{x} + 14 = 0$$
or
$$\left(- 3^{x + 2} + 9^{x} + 14\right) + 0 = 0$$
Do replacement
$$v = 3^{x}$$
we get
$$v^{2} - 9 v + 14 = 0$$
or
$$v^{2} - 9 v + 14 = 0$$
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
$$a = 1$$
$$b = -9$$
$$c = 14$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 14 + \left(-9\right)^{2} = 25$$
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
$$v_{1} = 7$$
Simplify
$$v_{2} = 2$$
Simplify
do backward replacement
$$3^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(3 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}$$
$$x_{2} = \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}} = \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}$$
Rapid solution
[src]
log(2)
x_1 = ------
log(3)
$$x_{1} = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}$$
log(7)
x_2 = ------
log(3)
$$x_{2} = \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}$$
Sum and product of roots
[src]
sum
log(2) log(7) ------ + ------ log(3) log(3)
$$\left(\frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}\right) + \left(\frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}\right)$$
=
log(2) log(7) ------ + ------ log(3) log(3)
$$\frac{\log{\left(2 \right)}}{\log{\left(3 \right)}} + \frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}$$
product
log(2) log(7) ------ * ------ log(3) log(3)
$$\left(\frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}\right) * \left(\frac{\log{\left(7 \right)}}{\log{\left(3 \right)}}\right)$$
=
log(2)*log(7)
-------------
2
log (3)
$$\frac{\log{\left(2 \right)} \log{\left(7 \right)}}{\log{\left(3 \right)}^{2}}$$
The graph