Solve the equation log(2) [x(x-1)]=1
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We have to solve the equation log(2) [x(x-1)]=1
As the base of the logarithm is 2 we can rewrite it as:
x(x-1) = 2^1
=> x^2 - x = 2
=> x^2 - x - 2 = 0
=> x^2 - 2x +x - 2 = 0
=> x(x - 2) +1(x - 2) = 0
=> (x +1)(x - 2) = 0
So we get x = 2 and x = -1, but as the logarithm of a negative number is not defined we eliminate x = -1
Therefore x = 2
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log(2) {x(x-2)} = 1.
We have to solve for x.
Solution:
We know that if log(a) m = n , then a^n = m.
Therefore if log(2) (x(x-1), then x(x-1) = 2^1 .
=> x^2- x = 2
=> x^2-x - 2 = 0
=> (x+1)(x-2) = 0
=> x+1 = 0, or x-2 = 0.
So x= -1, or x= 2.
First, we'll impose the constraints of existence of logarithm:
x(x-1)>0
The range of admissible values for x is (1 ; +infinite).
We'll solve the equation by taking anti-logarithm:
x(x-1) = 2^1
We'll remove the brackets:
x^2 - x - 2 = 0
We'll apply quadratic formula:
x1 = [1 + sqrt(1 + 8)]/2
x1 = (1+3)/2
x1 = 2
x2 = -1
Since the second value of x doesn't belong to (-1; +infinite), we'll reject it.
We'll accept as solution of equation x = 2.
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