Question

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

Answer 1

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`