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

Question

5751 views around the world

You can reuse this answer
Creative Commons License

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 $dy=1/2(0.12)=0.024$

$ln(128/25)=ln(125/25)+ Delta y$

$ln(128/25) ~~ ln(125/25)+ dy~~1.609+0.024=1.633$