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Math Help - Natural Exponential Function

  1. #1
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    Natural Exponential Function

    Find the equation of the tangent to the curve defined by <br />
y=x-e^{-x} that is parallel to the line represented by 3x-y-9=0

    This is my work, can someone check it please? I'm not sure if i have the rules of the natural exponential function correct.

    First off I want to say, I know that e^{lnm} =m
    BUT

    is there anything for e^{1/lnm}=??

    Now I will continue.
    y=x-e^{-x}
    y'=1+e^{-x}

    y'=3 at x=?

    3=1+e^{-x}
    ln2=lne^{-x}
    ln2=-xlne
    x=-lne/ln2
    x=-1/ln2

    y=mx+b
    y=(-1/ln2)-e^{-(-1/ln2)}
    y=(-1/ln2)-e^{(1/ln2)}

    y=mx+b

    3(-1/ln2)+b= (-1/ln2)-e^{(1/ln2)}
    b= (2/ln2)-e^{(1/ln2)}


    Final answer:
    y=3x+ (2/ln2)-e^{(1/ln2)}
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Find the equation of the tangent to the curve defined by <br />
y=x-e^{-x} that is parallel to the line represented by 3x-y-9=0
    y = x - e^{-x}<br />

    y' = 1 + e^{-x}

    1 + e^{-x} = 3

    e^{-x} = 2

    -x = \ln(2)

    x = -\ln(2)

    note that -\ln(2) \ne \frac{1}{\ln(2)} ,
    -\ln(2) = \ln(2)^{-1} = \ln\left(\frac{1}{2}\right)


    to get the equation of the tangent line ...

    at x = -\ln(2)

    y = -\ln(2) - e^{\ln(2)} = -[\ln(2) + 2]

    y + [\ln(2) + 2] = 3[x + \ln(2)]

    y = 3x + 2[\ln(2) - 1]
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  3. #3
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    It might be easier if you use y = m(x-x0) +y0 instead of y = mx + b

    no you can't simplify e^(1/ln(m))
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  4. #4
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    Quote Originally Posted by skeeter View Post
    y = x - e^{-x}<br />

    y' = 1 + e^{-x}

    1 + e^{-x} = 3

    e^{-x} = 2

    -x = \ln(2)

    x = -\ln(2)

    note that -\ln(2) \ne \frac{1}{\ln(2)} ,
    -\ln(2) = \ln(2)^{-1} = \ln\left(\frac{1}{2}\right)


    to get the equation of the tangent line ...

    at x = -\ln(2)

    y = -\ln(2) - e^{\ln(2)} = -[\ln(2) + 2]

    y + [\ln(2) + 2] = 3[x + \ln(2)]

    y = 3x + 2[\ln(2) - 1]
    I'm not too sure of your second last line.. is there another way of putting it?
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  5. #5
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    Quote Originally Posted by skeske1234 View Post
    I'm not too sure of your second last line.. is there another way of putting it?
    y - y_1 = m(x - x_1)
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  6. #6
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    Quote Originally Posted by skeeter View Post
    y - y_1 = m(x - x_1)
    ok.. why do we have to use that here? if I was to use y=mx+b (my teacher uses this in class) how would it look? I am trying to figure it out using that way
    my work:

    y=mx+b
    -(ln2+2)=3(-ln2)+b
    -ln2-2=-3ln2+b
    -ln2+3ln2-2=b
    -ln2+ln8-2=b
    ln6-2=b

    edit: if this is still wrong, can you please show me using the y=mx+b formula? thanks
    Last edited by skeske1234; August 14th 2009 at 10:01 AM.
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  7. #7
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    sorry, still seem to be having some trouble with this one>>
    my result ends up to be....

    y=3x+ln6-2

    is this still wrong?
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  8. #8
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    Quote Originally Posted by skeske1234 View Post
    ok.. why do we have to use that here? if I was to use y=mx+b (my teacher uses this in class) how would it look? I am trying to figure it out using that way
    my work:

    y=mx+b
    -(ln2+2)=3(-ln2)+b
    -ln2-2=-3ln2+b
    -ln2+3ln2-2=b
    -ln2+ln8-2=b
    ln6-2=b

    edit: if this is still wrong, can you please show me using the y=mx+b formula? thanks
    you need to review your algebra for logs ...

    \ln(8) - \ln(2) \ne \ln(6)

    \ln(8) - \ln(2) = \ln\left(\frac{8}{2}\right) = \ln(4)
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