You might try $\ln(17)=\ln(16)+\int_{16}^{17} \frac1x dx$.
You can then estimate the integral as a rectangular area with width $1$ and height $\frac{1}{16.5}$.
This gives $\ln(17)\approx \ln(16)+\frac{2}{33}$
According to my calculator, $\ln(17)\approx 2.83321334$, and $\ln(16)+\frac{2}{33}\approx 2.833194783$.
For better accuracy, you could partition $[16,17]$ into two (or more) subintervals.