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

1 Answer
Aug 25, 2014

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.