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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- Oct 9th 2008, 10:49 AMskyslimitLinear 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. - Oct 9th 2008, 01:47 PMskeeter
use the equation of the tangent line to $\displaystyle y = \ln{x}$ at $\displaystyle x = 1$ ...

$\displaystyle y' = \frac{1}{x}$

$\displaystyle y'(1) = 1$

$\displaystyle y - 0 = 1(x - 1)$

$\displaystyle y = x - 1$

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