Posted by APPRECIATIVE STUDENT on Wednesday, January 19, 2011 at 8:57am.
Thank you for your help.
I don't know how to do this, so if you could explain step by step how to do this, that would be really good....
"A rough approximation for ln(5) is 1.609. Use this apporximation and differentials to approximate ln(521/100)
thanks

math  Reiny, Wednesday, January 19, 2011 at 12:03pm
let y = ln(x)
x = 5
dy/dx = 1/x
dy = (1/x)dx
dx = 521/100  5 = .21 , y = ln(5) = 1.609
dy = (1/5)(.21) = .042
ln(521/100)
= ln (5.21) = appr. y + dy
= 1.609 + .042 = 1.651
check: ln(5.21) = 1.6505 (by calculator)
not bad