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x≥0 ∧ x≠1
4x-3≥0
4x≥3
x≥¾
suma sumarum: x ∈ <¾, ∞) \ {1}
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log(x) 4x-3 = 2
x² = 4x-3
x²-4x+3 = 0
Δ = 16-4*1*3 = 4
x₁ = (4-2)/2 = 1 ∉ Df
x₂ = (4+2)/2 = 3
Odp: x=3