Hi 2log(base 2) x=3+log(base 2) (x+16 2^3 = x^2/(x+16) 8(x+16) = x^2 x^2 -8x - 128 = 0 either by factoring or using the Quadratic formula (x+8) = 0 x = -8 this is an extraneous solution (x-16) = 0 x = 16