Express the given quantity as a single logarithm?

2015-09-04 5:31 am
(1/3)(ln(x+2)^3)+(1/2)[lnx - ln(x^2 + 3x +2)^2]

回答 (3)

2015-09-04 6:47 am
Please read :
2015-09-04 5:37 am
you have ln ( x + 2 ) + ln x^(1/2) - ln (x² + 3x + 2 ) = ln [ ( x^(3/2) + 2 √x ) / ( x² + 3x + 2 ) ]
2015-09-04 7:19 am
= (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)] }


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