Question

How do you use the tangent line approximation to approximate the value of `ln(1003)` ?

Answer 1

The answer is `3ln(10)+.003`

Another term for tangent line approximation is linear approximation. The linear approximation function is:

`L(x)~~f(a)+f'(a)(x-a)`

So we need to find the derivative:

`f(x)=ln(x)`
`f'(x)=1/x`

Now, we need to pick an `a` that is easy to compute. Unfortunately, with `ln` there isn't an easy way to compute a decimal value. So we will go with `a=1000=10^3`:

`f(a)=f(10^3)=ln(10^3)=3ln(10)`
`f'(a)=f'(1000)=1/(1000)`

So our linear approximation is:

`L(x)~~3ln(10)+1/(1000)(x-1000)`
`L(1003)~~3ln(10)+1/(1000)(1003-1000)`
`~~3ln(10)+3/(1000)`
`~~3ln(10)+.003`

We should leave this as the answer since it's supposed to be mental math. But let's look at how accurate this is: `ln(1003)=6.910750788` and `L(1003)~~6.910755279` which is accurate to 5 decimal places; so that's pretty good.