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**

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.**