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

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Answer 1 · 2015-03-22T14:35:41

To approximate #ln(128/25)# using linear approximation and/or differentials we need a number near #128/25# whose #ln# we know.

Clearly #128/25# is somewhat near #125/25#

and #125/25=5# whose #ln# we were given.

The difference between #ln(128/25)# and #ln(5)# is approximately equal to the differential of #y=lnx#

#dy=1/x dx#

To approximate near #5#, we use #dy = 1/5 dx= 1/5 (x-5)#

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

and #dy=1/2(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#