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    right i have dont most of the question and i also have the answer sheet but when i try to work out how they have done it, i am completely lost

    after integrating i got this with values for 'u' as '1/2 and 0'

    [ln|1 + u| − 3 ln|1 − u| + 2 ln|2 − u|]

    = (ln 3/2− 3 ln 1/2 + 2 ln 3/2) − (0 + 0 + 2 ln 2)
    = 3 ln 3/2 + 3 ln 2 − 2 ln 2
    = 3 ln 3 − 3 ln 2 + ln 2
    = 3 ln 3 − 2 ln 2

    i am lost after the second stage with simplifying them... its probably quite obvious but please could anyone clear this up for me??

    Thanks!!
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    = ln(3/2) - 3ln(1/2) + 2ln(3/2) - 2ln(2)
    = ln(3) - ln(2) - (3ln(1) - 3ln(2)) + 2ln(3) - 2ln(2) - 2ln(2)
    = ln(3) - ln(2) + - 0 + 3ln(2) + 2ln(3) - 2ln(2) - 2ln(2)
    = 3ln(3) - 2ln(2)

    then to simplify it further, continue using
    ln(a) - ln(b) = ln(a/b)

    and then at the end also use

    a ln(b) = ln(b^a)



    hope that makes some vague sort of sense
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    (Original post by lggodd)
    right i have dont most of the question and i also have the answer sheet but when i try to work out how they have done it, i am completely lost

    after integrating i got this with values for 'u' as '1/2 and 0'

    [ln|1 + u| − 3 ln|1 − u| + 2 ln|2 − u|]

    = (ln 3/2− 3 ln 1/2 + 2 ln 3/2) − (0 + 0 + 2 ln 2)
    = 3 ln 3/2 + 3 ln 2 − 2 ln 2
    = 3 ln 3 − 3 ln 2 + ln 2
    = 3 ln 3 − 2 ln 2

    i am lost after the second stage with simplifying them... its probably quite obvious but please could anyone clear this up for me??

    Thanks!!
    3\ln3=\ln27

    2\ln2=\ln4

    Put them together to get a single log.
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    C4 and logs?
    You're not planing on blowing up the trees are ye?
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    Thanks everyone, it has cleared up how they simplified it!

    Thank god for that, I'm getting some sleep before the exam I think :P
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    (Original post by Fatal_Microbes)
    C4 and logs?
    You're not planing on blowing up the trees are ye?

    haha i wish i could just blow up Core4 Maths in genaral...
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    (Original post by didgeridoo12uk)
    = ln(3/2) - 3ln(1/2) + 2ln(3/2) - 2ln(2)
    = ln(3) - ln(2) - (3ln(1) - 3ln(2)) + 2ln(3) - 2ln(2) - 2ln(2)
    = ln(3) - ln(2) + - 0 + 3ln(2) + 2ln(3) - 2ln(2) - 2ln(2)
    = 3ln(3) - 2ln(2)

    then to simplify it further, continue using
    ln(a) - ln(b) = ln(a/b)

    and then at the end also use

    a ln(b) = ln(b^a)



    hope that makes some vague sort of sense
    awesome help thanks
 
 
 
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