Răspuns:
Explicație pas cu pas:
1) log₃(3x-1) = log₃(2x+1)
Conditii : x > 1/3 ; x > -1/2 => x ∈ (1/3 ; +∞)
3x-1 = 2x+1 => x = 2
2) log₂(2x+5) = log₂(x²+3x+3)
Conditii : x > -5/2 ; x²+3x+3 > 0 ∀ x ∈ R (Δ < 0) =>
x ∈ (-5/2 ; +∞)
2x+5 = x²+3x+3 => x²+x-2 = 0 =>
x₁,₂ = [-1±√(1+8)]/2 = (-1±3)/2 =>
x₁ = -2 ; x₂ = 1 ambele sunt solutii
3) log₅(x²-4) = log₅(x+4)
Conditii :
x²-4 > 0 => x ∈ (-∞ ; -2)∪ (2 ; +∞)
x+4 > 0 => x > -4 =>
x ∈ (-4 ; -2) ∪ (2 ; +∞)
x²-4 = x+4 => x²-x-8 = 0 =>
x₁,₂ = (1±√33)/2
