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Thread: linear approximation... of ln(1.01)

  1. #1
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    linear approximation... of ln(1.01)

    i pretty much understand linear approximations but i cant seem to solve this problem. if anyone can show me some steps to get me started i would really love that
    Use linear approximation to approximate the number ln(1.01)

    we know that
    ln1 = 0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by delcidm7 View Post
    i pretty much understand linear approximations but i cant seem to solve this problem. if anyone can show me some steps to get me started i would really love that
    Use linear approximation to approximate the number ln(1.01)

    we know that
    ln1 = 0
    yes, we do know that

    let $\displaystyle f(x) = \ln x$

    use the linear approximation formula: $\displaystyle f(x) \approx f(a) + f'(a)(x - a)$

    here, x = 1.01, a = 1
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  3. #3
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    ok so i got
    1st: $\displaystyle f(1.01) \approx f(1) + f'(1)(1.01 - 1)$
    then: $\displaystyle f(1.01) \approx 0 + 1(.01)$
    finally: $\displaystyle f(1.01) \approx (.01)$
    thanks so much!!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by delcidm7 View Post
    ok so i got

    $\displaystyle
    f(1.01) \approx f(1) + f'(1)(1.01 - 1)
    $


    $\displaystyle
    f(1.01) \approx 0 + 1(.01)$

    $\displaystyle
    f(1.01) \approx (.01)$

    thanks so much!!
    correct
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