To approximate ln(128/25)ln(12825) using linear approximation and/or differentials we need a number near 128/2512825 whose lnln we know.
Clearly 128/2512825 is somewhat near 125/2512525
and 125/25=512525=5 whose lnln we were given.
The difference between ln(128/25)ln(12825) and ln(5)ln(5) is approximately equal to the differential of y=lnxy=lnx
dy=1/x dxdy=1xdx
To approximate near 55, we use dy = 1/5 dx= 1/5 (x-5)dy=15dx=15(x−5)
With x=128/25x=12825, x-5=3/25=12/100=0.12x−5=325=12100=0.12
and dy=1/2(0.12)=0.024dy=12(0.12)=0.024
ln(128/25)=ln(125/25)+ Delta y
ln(128/25) ~~ ln(125/25)+ dy~~1.609+0.024=1.633