Linear Approximation Question

• Oct 9th 2008, 11:49 AM
skyslimit
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.
• Oct 9th 2008, 02:47 PM
skeeter
use the equation of the tangent line to $y = \ln{x}$ at $x = 1$ ...

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

$y'(1) = 1$

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

$y = x - 1$

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