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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 x^4=(x-56)^2
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Identical expressions
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- 20*5^(2*x)-12*5^x+1=O
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Similar expressions
- 20*5^(2*x)+12*5^x+1=0
- 20*5^(2*x)-12*5^x-1=0
20*5^(2*x)-12*5^x+1=0 equation
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The solution
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[src]
2*x x 20*5 - 12*5 + 1 = 0
$$\left(20 \cdot 5^{2 x} - 12 \cdot 5^{x}\right) + 1 = 0$$
Detail solution
Given the equation:
$$\left(20 \cdot 5^{2 x} - 12 \cdot 5^{x}\right) + 1 = 0$$
or
$$\left(20 \cdot 5^{2 x} - 12 \cdot 5^{x}\right) + 1 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$20 v^{2} - 12 v + 1 = 0$$
or
$$20 v^{2} - 12 v + 1 = 0$$
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
$$a = 20$$
$$b = -12$$
$$c = 1$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = \frac{1}{2}$$
$$v_{2} = \frac{1}{10}$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(5 \right)}} = - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
$$x_{2} = \frac{\log{\left(\frac{1}{10} \right)}}{\log{\left(5 \right)}} = -1 - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
$$\left(20 \cdot 5^{2 x} - 12 \cdot 5^{x}\right) + 1 = 0$$
or
$$\left(20 \cdot 5^{2 x} - 12 \cdot 5^{x}\right) + 1 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$20 v^{2} - 12 v + 1 = 0$$
or
$$20 v^{2} - 12 v + 1 = 0$$
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
$$a = 20$$
$$b = -12$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(-12)^2 - 4 * (20) * (1) = 64
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = \frac{1}{2}$$
$$v_{2} = \frac{1}{10}$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{1}{2} \right)}}{\log{\left(5 \right)}} = - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
$$x_{2} = \frac{\log{\left(\frac{1}{10} \right)}}{\log{\left(5 \right)}} = -1 - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
Rapid solution
[src]
-log(2)
x1 = --------
log(5)
$$x_{1} = - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
-log(10)
x2 = ---------
log(5)
$$x_{2} = - \frac{\log{\left(10 \right)}}{\log{\left(5 \right)}}$$
x2 = -log(10)/log(5)
Sum and product of roots
[src]
sum
log(2) log(10) - ------ - ------- log(5) log(5)
$$- \frac{\log{\left(10 \right)}}{\log{\left(5 \right)}} - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
=
log(2) log(10) - ------ - ------- log(5) log(5)
$$- \frac{\log{\left(10 \right)}}{\log{\left(5 \right)}} - \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}}$$
product
-log(2) -log(10) --------*--------- log(5) log(5)
$$- \frac{\log{\left(2 \right)}}{\log{\left(5 \right)}} \left(- \frac{\log{\left(10 \right)}}{\log{\left(5 \right)}}\right)$$
=
log(2)*log(10)
--------------
2
log (5)
$$\frac{\log{\left(2 \right)} \log{\left(10 \right)}}{\log{\left(5 \right)}^{2}}$$
log(2)*log(10)/log(5)^2