Thank you for your help.

Question

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

Answer 1

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