Question

How do you use the tangent line approximation to approximate the value of `ln(1005)` ?

Answer 1

If you meant to approximate `ln(1.005)`, then we can do the following.

`ln(1.005) approx 0.005`

Let us look at some details.

Let `f(x)=lnx`. `Rightarrow f'(x)=1/x`

`f(1)=ln1=0`
`f'(1)=1/1=1`

So, the tangent line approximation `L(x)` can be written as

`L(x)=f(1)+f'(1)(x-1)=0+1(x-1)=x-1`

Hence,

`f(1.005) approx L(1.005)=1.005-1=0.005`

I hope that this was what you were looking for.