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

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Answer 1 · 2015-07-19T13:46:17

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

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

Take the exponential base $2$ of both sides:

$x^2-2x = 2^(log_2(x^2-2x)) = 2^3 = 8$

Subtract $8$ from both ends to get:

$0 = x^2-2x-8 = (x-4)(x+2)$

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

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