Express the given quantity as a single logarithm?
(1/3)(ln(x+2)^3)+(1/2)[lnx - ln(x^2 + 3x +2)^2]
回答 (3)
you have ln ( x + 2 ) + ln x^(1/2) - ln (x² + 3x + 2 ) = ln [ ( x^(3/2) + 2 √x ) / ( x² + 3x + 2 ) ]
= (1/3).Ln(x + 2)³ + (1/2).[Ln(x) - Ln(x² + 3x + 2)²]
= (1/3).Ln(x + 2)³ + (1/2).Ln(x) - (1/2).Ln(x² + 3x + 2)² → recall: Ln(x^a) = a.Ln(x)
= [(1/3) * 3.Ln(x + 2)] + (1/2).Ln(x) - [(1/2) * 2.Ln(x² + 3x + 2)]
= Ln(x + 2) + (1/2).Ln(x) - Ln(x² + 3x + 2) → you know that: a.Ln(x) = Ln(x^a)
= Ln(x + 2) + Ln(x)^(1/2) - Ln(x² + 3x + 2) → you know that: a^(1/2) = √a
= Ln(x + 2) + Ln(√x) - Ln(x² + 3x + 2) → recall: Ln(a) + Ln(b) = Ln(ab)
= Ln[(x + 2).√x] - Ln(x² + 3x + 2) → recall: Ln(a) - Ln(b) = Ln(a/b)
= Ln{ [(x + 2).√x] / (x² + 3x + 2) } → you know that: (x² + 3x + 2) = (x + 1).(x + 2)
= Ln{ [(x + 2).√x] / [(x + 1).(x + 2)] } → you can simplify by (x + 2)
= Ln{ √x / (x + 1)] }
收錄日期: 2021-04-18 00:13:08
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