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Math Help - Tangent line approximation

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    Tangent line approximation

    My problem is ln(55). My answer was 8/5e^4. My work is as follows f(x) = ln x
    f'(x)= 1/x
    55 appx = e^4 which is 54.6
    f(e^4)= 4
    f'(e^4)= 4/(e^4)
    x-a=h= 55 - 54.6 = .4

    f(55)= f(55+(55-54.6)) appx = f(e^4) + f'(e^4)*(55 - 54.6)
    appx = 4 + 4/(e^4) (55-54.6) = 4 + (8/(e^4)) - 4 = 8/5(e^4)
    When you type this in the calculator ln(55) = 4.007 and 8/(5(e^4)) = .029 which is not close at all. Can anyone explain to me what I'm doing wrong, or am missing?
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  2. #2
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    Quote Originally Posted by spazzyskylar View Post
    My problem is ln(55). My answer was 8/5e^4. My work is as follows f(x) = ln x
    f'(x)= 1/x
    55 appx = e^4 which is 54.6
    f(e^4)= 4
    f'(e^4)= 4/(e^4)
    x-a=h= 55 - 54.6 = .4

    f(55)= f(55+(55-54.6)) appx = f(e^4) + f'(e^4)*(55 - 54.6)
    appx = 4 + 4/(e^4) (55-54.6) = 4 + (8/(e^4)) - 4 = 8/5(e^4)
    When you type this in the calculator ln(55) = 4.007 and 8/(5(e^4)) = .029 which is not close at all. Can anyone explain to me what I'm doing wrong, or am missing?
    f(x) = \ln{x}

    f'(x) = \frac{1}{x}

    let x = e^4 , y = 4

    tangent line equation at the point (e^4,4) ...

    y - 4 = \frac{1}{e^4}(x - e^4)

    y - 4 = \frac{x}{e^4} - 1

    y = \frac{x}{e^4} + 3

    f(55) \approx  \frac{55}{e^4} + 3 = 4.00736013888...

    note that \ln(55) = 4.00733318523...

    not a bad approximation
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