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uer23
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#1
Report Thread starter 8 years ago
#1
Hi!

This is part of a C4 integration problem. I'm having problem from here onwards...

(9 - 2\ln 3) - (\frac{8}{3} - 2\ln 2)

9 - 2\ln 3 - \frac{8}{3} + 2\ln 2

\frac{19}{3} + 2\ln 3^{-1} + 2\ln 2

\frac{19}{3} + 2\ln \frac{1}{3} + 2\ln 2

\frac{19}{3} + 2\ln \frac{2}{3}

The answer is supposed to be :

\frac{19}{3} - 2\ln \frac{3}{2}

Can someone tell me where I'm going wrong. Thanks.
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abbii
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#2
Report 8 years ago
#2
19/3 - 2ln3 + 2ln2
19/3 - (2ln3 - 2ln2)
19/3 - 2(ln3 - ln2)
19/3 - 2(ln(3/2))
19/3 - 2ln(3/2)
thats how you get the answer, cant see where youve gone wrong though
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Sasukekun
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#3
Report 8 years ago
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Well technically both are correct, I'm just not sure how they got to theirs.

2ln(2/3) = -2ln(3/2)

-2ln(3/2) = 2ln((3/2)^-1)

= +2ln(2/3)
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Physics Enemy
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#4
Report 8 years ago
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Both answers are right.

19 + 2ln(2/3) = 19 + 2ln[(3/2)^-1] = 19 - 2ln(3/2)
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