# How do you solve log_2(x^2 - 2x) =3?

Jul 19, 2015

Take the exponential base $2$ of both sides, then solve the resulting quadratic to find:

$x = 4$ or $x = - 2$

#### Explanation:

Take the exponential base $2$ of both sides:

${x}^{2} - 2 x = {2}^{{\log}_{2} \left({x}^{2} - 2 x\right)} = {2}^{3} = 8$

Subtract $8$ from both ends to get:

$0 = {x}^{2} - 2 x - 8 = \left(x - 4\right) \left(x + 2\right)$

So $x = 4$ or $x = - 2$