Use linear approximation to approximate the number $ln(1.02)$.

This is what I did and it is still wrong on my online homework.

$f(x) = ln(x)$

$f'(x) = \dfrac{1}{x}$


$y=\dfrac{1}{1.02}(1.02-1)= 0.0196$

  • 1
    $\begingroup$ You can just give me the reasoning on what I did wrong and I will try to work it out. I dont want you to think I want the answer. Just that I am better working backwards with the answer. $\endgroup$
    – Zealotory
    Nov 14 '14 at 9:03

You want to use $$f(x) \approx f(a) + f^{\prime}(a) (x-a)$$

What you wrote down is $$f(x) \approx f(a) + f^{\prime}(\pmb{x})(x-a)$$

  • $\begingroup$ Oh I think I get it now! Thanks $\endgroup$
    – Zealotory
    Nov 14 '14 at 9:09
  • $\begingroup$ If you do it correctly, you should simply get 0.02. $\endgroup$
    – JessicaK
    Nov 14 '14 at 9:10

The idea is to use $$\ln(x+h)\sim \ln(x)+(\ln)'(x)h$$

In our case this would be $$\ln(1+0.02)\sim \ln(1)+\frac{1}{1}(0.02)$$

using that $(\ln)'(x)=\frac{1}{x}$.


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