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.
use the equation of the tangent line to $\displaystyle y = \ln{x}$ at $\displaystyle x = 1$ ...
$\displaystyle y' = \frac{1}{x}$
$\displaystyle y'(1) = 1$
$\displaystyle y - 0 = 1(x - 1)$
$\displaystyle y = x - 1$
$\displaystyle \ln(1.38) \approx 1.38 - 1 = 0.38$