Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation exp(x)=0
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Equation 2*x^2+5*x-25=13*x+17
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Equation 3*x-15=0
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Equation 0=2-3x
- Express {x} in terms of y where:
- 9*x-19*y=-20
- 15*x+17*y=-10
- -11*x-18*y=6
- -7*x-5*y=-17
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Identical expressions
- twenty-five ^x-(a+ one)* five ^x+3a- six = zero
- 25 to the power of x minus (a plus 1) multiply by 5 to the power of x plus 3a minus 6 equally 0
- twenty minus five to the power of x minus (a plus one) multiply by five to the power of x plus 3a minus six equally zero
- 25x-(a+1)*5x+3a-6=0
- 25x-a+1*5x+3a-6=0
- 25^x-(a+1)5^x+3a-6=0
- 25x-(a+1)5x+3a-6=0
- 25x-a+15x+3a-6=0
- 25^x-a+15^x+3a-6=0
- 25^x-(a+1)*5^x+3a-6=O
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Similar expressions
- 25^x-(a+1)*5^x-3a-6=0
- 25^x-(a-1)*5^x+3a-6=0
- 25^x-(a+1)*5^x+3a+6=0
- 25^x+(a+1)*5^x+3a-6=0
25^x-(a+1)*5^x+3a-6=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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x x 25 - (a + 1)*5 + 3*a - 6 = 0
$$\left(3 a + \left(25^{x} - 5^{x} \left(a + 1\right)\right)\right) - 6 = 0$$
Detail solution
Given the equation:
$$\left(3 a + \left(25^{x} - 5^{x} \left(a + 1\right)\right)\right) - 6 = 0$$
or
$$\left(3 a + \left(25^{x} - 5^{x} \left(a + 1\right)\right)\right) - 6 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
or
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
Expand the expression in the equation
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
We get the quadratic equation
$$- a v + 3 a + v^{2} - v - 6 = 0$$
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
$$a = 1$$
$$b = - a - 1$$
$$c = 3 a - 6$$
, then
The equation has two roots.
or
$$v_{1} = \frac{a}{2} + \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2}$$
$$v_{2} = \frac{a}{2} - \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2}$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{a}{2} + \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}} = \frac{\log{\left(\frac{a}{2} + \frac{\sqrt{a^{2} - 10 a + 25}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}}$$
$$x_{2} = \frac{\log{\left(\frac{a}{2} - \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}} = \frac{\log{\left(\frac{a}{2} - \frac{\sqrt{a^{2} - 10 a + 25}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}}$$
$$\left(3 a + \left(25^{x} - 5^{x} \left(a + 1\right)\right)\right) - 6 = 0$$
or
$$\left(3 a + \left(25^{x} - 5^{x} \left(a + 1\right)\right)\right) - 6 = 0$$
Do replacement
$$v = 5^{x}$$
we get
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
or
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
Expand the expression in the equation
$$3 a + v^{2} + v \left(- a - 1\right) - 6 = 0$$
We get the quadratic equation
$$- a v + 3 a + v^{2} - v - 6 = 0$$
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
$$a = 1$$
$$b = - a - 1$$
$$c = 3 a - 6$$
, then
D = b^2 - 4 * a * c =
(-1 - a)^2 - 4 * (1) * (-6 + 3*a) = 24 + (-1 - a)^2 - 12*a
The equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = \frac{a}{2} + \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2}$$
$$v_{2} = \frac{a}{2} - \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2}$$
do backward replacement
$$5^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(\frac{a}{2} + \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}} = \frac{\log{\left(\frac{a}{2} + \frac{\sqrt{a^{2} - 10 a + 25}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}}$$
$$x_{2} = \frac{\log{\left(\frac{a}{2} - \frac{\sqrt{- 12 a + \left(- a - 1\right)^{2} + 24}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}} = \frac{\log{\left(\frac{a}{2} - \frac{\sqrt{a^{2} - 10 a + 25}}{2} + \frac{1}{2} \right)}}{\log{\left(5 \right)}}$$
The graph
Rapid solution
[src]
log(3)
x1 = ------
log(5)
$$x_{1} = \frac{\log{\left(3 \right)}}{\log{\left(5 \right)}}$$
log(|-2 + a|) I*arg(-2 + a)
x2 = ------------- + -------------
log(5) log(5)
$$x_{2} = \frac{\log{\left(\left|{a - 2}\right| \right)}}{\log{\left(5 \right)}} + \frac{i \arg{\left(a - 2 \right)}}{\log{\left(5 \right)}}$$
x2 = log(|a - 2|)/log(5) + i*arg(a - 2)/log(5)
Sum and product of roots
[src]
sum
log(3) log(|-2 + a|) I*arg(-2 + a) ------ + ------------- + ------------- log(5) log(5) log(5)
$$\left(\frac{\log{\left(\left|{a - 2}\right| \right)}}{\log{\left(5 \right)}} + \frac{i \arg{\left(a - 2 \right)}}{\log{\left(5 \right)}}\right) + \frac{\log{\left(3 \right)}}{\log{\left(5 \right)}}$$
=
log(3) log(|-2 + a|) I*arg(-2 + a) ------ + ------------- + ------------- log(5) log(5) log(5)
$$\frac{\log{\left(\left|{a - 2}\right| \right)}}{\log{\left(5 \right)}} + \frac{i \arg{\left(a - 2 \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(3 \right)}}{\log{\left(5 \right)}}$$
product
log(3) /log(|-2 + a|) I*arg(-2 + a)\ ------*|------------- + -------------| log(5) \ log(5) log(5) /
$$\frac{\log{\left(3 \right)}}{\log{\left(5 \right)}} \left(\frac{\log{\left(\left|{a - 2}\right| \right)}}{\log{\left(5 \right)}} + \frac{i \arg{\left(a - 2 \right)}}{\log{\left(5 \right)}}\right)$$
=
(I*arg(-2 + a) + log(|-2 + a|))*log(3)
--------------------------------------
2
log (5)
$$\frac{\left(\log{\left(\left|{a - 2}\right| \right)} + i \arg{\left(a - 2 \right)}\right) \log{\left(3 \right)}}{\log{\left(5 \right)}^{2}}$$
(i*arg(-2 + a) + log(|-2 + a|))*log(3)/log(5)^2