I believe you meant to write [tex]log_{2}(x+2) + log_{2}(x-2) \leq log_{2}(5) [/tex] ? If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always! Remember: log a + log b = log (a*b) So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4) Now back to the inequality: log (x^2 - 4) ≤ log 5 Raise both sides as powers of 2 ( Since it's the base of your log) Now, x^2 - 4 ≤ 5 Add 4 both sides: x^2 ≤ 9 Square root both sides x ≤ +3 or x ≤ -3 Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets. So x ≤ 3 But can never be less than 2 as well for the same previous reason. Hope that helped.
