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C3 - OCR paul Sanders book - sample paper at the back watch

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    6) two graphs of y = 3e^-x and y = 2e^3x-1

    the graphs intersect at the point whose x is P

    prove that P = 1/4ln(3e/2)

    any ideas how to do this?
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    When the two graphs intersect

    3e^{-x} = 2e^{3x} - 1

    Solve for x.
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    lol i know that but than what do i do next
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    3/2 e^(-x)= e^(3x-1)

    Multiply through by e^x:
    3/2=e^4x-1

    Multiply through by e:
    3e/2=e^4x

    Ln both sides:
    4x=Ln3e/2
    x=(1/4)ln(3e/2)

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    Well i was trying to get you to think about that...
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    aaw thanks Ben i worked it out some longgg way using law of logs.
    i ln -ed both sides

    ln(2e^3x-1)=ln(3e^-x)
    ln2 + 3x-1 = ln3 -x
    4x = ln3 - ln2 + 1
    4x = ln(3/2) + ln(e) note :- ln(e) =1
    4x = ln(3e/2)
    and than i finally got P = 1/4ln(3e/2)

    yayy loll thanks BenSpurgen for your version though definately much easier. thought id post my way just in case anyone may be stuck on the same thing.
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    No problems
 
 
 
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Updated: January 11, 2008

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