ln(x+1) + ln(x-1) = 2 D = ] 1 ; ∞ [ [ Argumente des ln > 0 ! ]
⇔ ln( (x+1) • (x-1) ) = 2 [ Logarithmensatz log(a • b) = log(a) + log(b) ]
⇔ (x+1) • (x-1) = e2
⇔ x2 - 1 = e2
⇔ x2 = e2 +1
⇔ x = √(e2 + 1) ≈ 2.896386731 (die negative Lösung entfällt, da ∉ D)
Gruß Wolfgang