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Thread: Linear Approximation Question

  1. October 9th 2008, 10:49 AM #1
    skyslimit
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    Linear Approximation Question

    How can I find the linear approximation of ln(1.38) when x=1

    Natural Log questions always confuse me. Is ther anything I can do to help my understanding of these types of problems so I can shoot them out with ease.
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  2. October 9th 2008, 01:47 PM #2
    skeeter
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    use the equation of the tangent line to y = \ln{x} at x = 1 ...

    y' = \frac{1}{x}

    y'(1) = 1

    y - 0 = 1(x - 1)

    y = x - 1

    \ln(1.38) \approx 1.38 - 1 = 0.38
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