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

Aug 25, 2014

The answer is $3 \ln \left(10\right) + .003$

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

$L \left(x\right) \approx f \left(a\right) + f ' \left(a\right) \left(x - a\right)$

So we need to find the derivative:

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

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 $a = 1000 = {10}^{3}$:

$f \left(a\right) = f \left({10}^{3}\right) = \ln \left({10}^{3}\right) = 3 \ln \left(10\right)$
$f ' \left(a\right) = f ' \left(1000\right) = \frac{1}{1000}$

So our linear approximation is:

$L \left(x\right) \approx 3 \ln \left(10\right) + \frac{1}{1000} \left(x - 1000\right)$
$L \left(1003\right) \approx 3 \ln \left(10\right) + \frac{1}{1000} \left(1003 - 1000\right)$
$\approx 3 \ln \left(10\right) + \frac{3}{1000}$
$\approx 3 \ln \left(10\right) + .003$

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