C.E.
{x - 2 > 0
{x^2 - 4 > 0
{x + 2 > 0
risolvo: [x > 2]
2·(- LN(x - 2)/(2·LN(5))) + LN(x^2 - 4)/(2·LN(5)) = LN(x + 2)/LN(5)
(cambio base)
(2·(- LN(x - 2)/(2·LN(5))) + LN(x^2 - 4)/(2·LN(5)) = LN(x + 2)/LN(5))·2·LN(5)
LN(x^2 - 4) - 2·LN(x - 2) = 2·LN(x + 2)
LN((x^2 - 4)/(x - 2)^2) = LN((x + 2)^2)
(x^2 - 4)/(x - 2)^2 = (x + 2)^2
(x + 2)/(x - 2) = (x + 2)^2
x + 2 = (x + 2)^2·(x - 2)
(x + 2)^2·(x - 2) - (x + 2) = 0
(x + 2)·(x^2 - 5) = 0
x = - √5 ∨ x = √5 ∨ x = -2
In grassetto l'unica accettabile: x = √5