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    (Original post by Robbie242)
    The x-coordinate is correct

    Let me find the y-coordinate

    when x=\dfrac{1}{e}, y=\dfrac{1}{e}\ln(\dfrac{1}{e} )

    Using log laws we have y=\dfrac{1}{e}(\ln(1)-\ln(e))

    \ln(e)=1 so y=-\dfrac{1}{e}

    I'm not sure about reasoning actually but working out the actual coordinate gives the answer the book has
    are you rewriting ln(1/e) as two logs ln(1) - ln(e) ?
    now sure how you get -1/e from that
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    (Original post by jacksonmeg)
    are you rewriting ln(1/e) as two logs ln(1) - ln(e) ?
    now sure how you get -1/e from that
    Yeah
    It's the division law

    \ln(a)-\ln(b)=\ln\left(\dfrac{a}{b} \right )

    In this case we have y=\dfrac{1}{e}\left( \ln(1)-\ln(e) \right)

    ln is the inverse of e so \ln(e)=1 and \ln(1)=0 (if you look at the graph)

    so y=\dfrac{1}{e}\left(0-1 \right) = -\dfrac{1}{e}
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    (Original post by Robbie242)
    Yeah
    It's the division law

    \ln(a)-\ln(b)=\ln\left(\dfrac{a}{b} \right )

    In this case we have y=\dfrac{1}{e}\left( \ln(1)-\ln(e) \right)

    ln is the inverse of e so \ln(e)=1 and \ln(1)=0 (if you look at the graph)

    so y=\dfrac{1}{e}\left(0-1 \right) = -\dfrac{1}{e}
    we havent actually done section 3 in lesson. we jumped straight to differentiation and integration, not sure why. but none of it makes any sense, i guess ill start on the questions on section 3
 
 
 
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