# How do you use the tangent line approximation to approximate the value of ln(1.006) ?

Sep 28, 2014

$\ln \left(1.006\right) \approx 0.006$

Let us look at some details.

Let $f \left(x\right) = \ln x$. $R i g h t a r r o w f ' \left(x\right) = \frac{1}{x}$

$f \left(1\right) = \ln 1 = 0$
$f ' \left(1\right) = \frac{1}{1} = 1$

So, the tangent line approximation $L \left(x\right)$ can be written as

$L \left(x\right) = f \left(1\right) + f ' \left(1\right) \left(x - 1\right) = 0 + 1 \left(x - 1\right) = x - 1$

Hence,

$f \left(1.006\right) \approx L \left(1.006\right) = 1.006 - 1 = 0.006$

I hope that this was helpful.