given eq. can be written as log₂ [(x+7)/(x+1)]² + log₂ [(x+1)/(x-1)] = 1

or log₂ [(x+7)²(x+1)/(x+1)²(x-1)] = 1

or log₂ [(x+7)²/(x²-1)] = 1

or (x+7)²/(x²-1) = 2

or x² + 14x +49 = 2x² - 2

or x²-14x -51 = 0

or x = -3, 17

since at x = -3 first part will become negative inside root so it will not be a root.

so only 1 root