Solve for x log base 2 of x^2 = log base 2 of 4x-4

$\log_{2} (x^{2}) = \log_{2} (4 x - 4)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} = 4 x - 4$

Step 2

Solve for $x$ .

Subtract $4 x$ from both sides of the equation.

$x^{2} - 4 x = - 4$

Add $4$ to both sides of the equation.

$x^{2} - 4 x + 4 = 0$

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$x^{2} - 4 x + 2^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

${(x - 2)}^{2} = 0$

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$