Mister Exam
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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
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- -5*x-6*y=-1
- -17*x-15*y=-17
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Identical expressions
- (twenty-five ^(x- one , five))-(twelve * five ^(x- two))+ seven = zero
- (25 to the power of (x minus 1,5)) minus (12 multiply by 5 to the power of (x minus 2)) plus 7 equally 0
- (twenty minus five to the power of (x minus one , five)) minus (twelve multiply by five to the power of (x minus two)) plus seven equally zero
- (25(x-1,5))-(12*5(x-2))+7=0
- 25x-1,5-12*5x-2+7=0
- (25^(x-1,5))-(125^(x-2))+7=0
- (25(x-1,5))-(125(x-2))+7=0
- 25x-1,5-125x-2+7=0
- 25^x-1,5-125^x-2+7=0
- (25^(x-1,5))-(12*5^(x-2))+7=O
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Similar expressions
- (25^(x-1,5))-(12*5^(x+2))+7=0
- (25^(x+1,5))-(12*5^(x-2))+7=0
- (25^(x-1,5))+(12*5^(x-2))+7=0
- (25^(x-1,5))-(12*5^(x-2))-7=0
(25^(x-1,5))-(12*5^(x-2))+7=0 equation
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The solution
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[src]
x - 3/2 x - 2 25 - 12*5 + 7 = 0
$$\left(25^{x - \frac{3}{2}} - 12 \cdot 5^{x - 2}\right) + 7 = 0$$
Detail solution
Given the equation:
$$\left(25^{x - \frac{3}{2}} - 12 \cdot 5^{x - 2}\right) + 7 = 0$$
or
$$\left(25^{x - \frac{3}{2}} - 12 \cdot 5^{x - 2}\right) + 7 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$\frac{v^{2}}{125} - \frac{12 v}{25} + 7 = 0$$
or
$$\frac{v^{2}}{125} - \frac{12 v}{25} + 7 = 0$$
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
$$a = \frac{1}{125}$$
$$b = - \frac{12}{25}$$
$$c = 7$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 35$$
$$v_{2} = 25$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(25 \right)}}{\log{\left(5 \right)}} = 2$$
$$x_{2} = \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}} = 1 + \frac{\log{\left(7 \right)}}{\log{\left(5 \right)}}$$
$$\left(25^{x - \frac{3}{2}} - 12 \cdot 5^{x - 2}\right) + 7 = 0$$
or
$$\left(25^{x - \frac{3}{2}} - 12 \cdot 5^{x - 2}\right) + 7 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$\frac{v^{2}}{125} - \frac{12 v}{25} + 7 = 0$$
or
$$\frac{v^{2}}{125} - \frac{12 v}{25} + 7 = 0$$
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
$$a = \frac{1}{125}$$
$$b = - \frac{12}{25}$$
$$c = 7$$
, then
D = b^2 - 4 * a * c =
(-12/25)^2 - 4 * (1/125) * (7) = 4/625
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 35$$
$$v_{2} = 25$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(25 \right)}}{\log{\left(5 \right)}} = 2$$
$$x_{2} = \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}} = 1 + \frac{\log{\left(7 \right)}}{\log{\left(5 \right)}}$$
Sum and product of roots
[src]
sum
log(35)
2 + -------
log(5)
$$2 + \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}}$$
=
log(35)
2 + -------
log(5)
$$2 + \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}}$$
product
log(35) 2*------- log(5)
$$2 \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}}$$
=
/ 2 \ | ------| | log(5)| log\35 /
$$\log{\left(35^{\frac{2}{\log{\left(5 \right)}}} \right)}$$
log(35^(2/log(5)))
Rapid solution
[src]
x1 = 2
$$x_{1} = 2$$
log(35)
x2 = -------
log(5)
$$x_{2} = \frac{\log{\left(35 \right)}}{\log{\left(5 \right)}}$$
x2 = log(35)/log(5)