Since we want ln(0.6), but make use of the series expansion for ln(1-x), x = 0.4

$\displaystyle \ln(1-x) = - \left(x + \frac{x^2}{2} + \frac{x^3}{3} + ... \right)$

$\displaystyle \ln(1-x) = - \sum ^{\infty} _{n=0} (-1) \frac{x^{n+1}}{n+1}

$

Let $\displaystyle e_n$ be the error estimate.

$\displaystyle | e_1 | = \frac{0.4^{1 + 1}}{1+1} = 0.08$

$\displaystyle | e_2 | = 0.02$

$\displaystyle | e_3 | = 0.0064$

So therefore $\displaystyle \ln(0.6) = \sum ^{2} _{n=0} (-1) \frac{0.4^{n+1}}{n+1} = -1,101333...$