Question

Tema la matematica ECUATII LOGARITMICE
1) log 3 (3x-1)= log3(2x+1)
b)log2(2x+5)= log2(x2+3+3)
c)log5(x2-4)= log5(x+4)

2) a) logx+2(x2-1)=logx+2(x+5)
b)logx+1(3x+2)=2

Answer 1

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