If a rough approximation for ln(5) is 1.609 how do you use this approximation and differentials to approximate ln(128/25)?

1 Answer
Mar 22, 2015

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(x5)

With x=128/25x=12825, x-5=3/25=12/100=0.12x5=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