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

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Answer 1 · 2014-09-28T23:05:35

#ln(1.006) approx 0.006#

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(1.006) approx L(1.006)=1.006-1=0.006#

I hope that this was helpful.