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

1 Answer
Jul 19, 2015

Answer:

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^2-2x = 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#