Answer
$\ln (\dfrac{5}{3}) \approx 0.510825624$
Work Step by Step
Consider the series for $\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-....+(-1)^{n-1}\dfrac{x^n}{n}$
As we are given that $x=\dfrac{2}{3}$
Now, $\ln (1+x)=\ln (1+\dfrac{2}{3})=\ln [\dfrac{(3+2)}{3}]$
Thus $\ln (\dfrac{5}{3}) \approx 0.510825624$
