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Posted by 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 - , 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

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