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

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Answer 1 · 2014-09-28T23:05:35

$ln (1.006) \approx 0.006$ $ln(1.006) approx 0.006$

Let us look at some details.

Let $f (x) = ln x$ $f(x)=lnx$ . $Rightarrow f'(x)=1/x$

$f(1)=ln1=0$
$f'(1)=1/1=1$

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.

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