Podstawa logarytmu musi być dodatnia, czyli > 0.
[tex]a) \ y = log_{2}(2x+8)\\\\2x+8 > 0 \ \ /:2\\\\x+4 > 0\\\\x > -4\\\\\boxed{x \in (-4; +\infty)}[/tex]
[tex]b) \ y = log_{4}(x^{2}-9)\\\\x^{2}-9 > 0\\\\M. \ zerowe:\\\\x^{2}-9 = 0\\\\(x+3)(x-3) = 0\\\\x+3 = 0 \ \vee \ x - 3 = 0\\\\x = -3 \ \vee \ x = 3\\\\a > 0, to \ ramiona \ paraboli \ skierowane \ do \ gory, \ wowczas:\\\\\boxed{x \in (-\infty;-3) \ \cup \ (3;+\infty) }[/tex]
