# Solve the equation log(2) [x(x-1)]=1

justaguide | Certified Educator

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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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neela | Student

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.

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giorgiana1976 | Student

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

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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