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April 24, 2014

April 24, 2014

Posted by **APPRECIATIVE STUDENT** on Wednesday, January 19, 2011 at 8:57am.

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 12:03pmlet 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

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