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Math Help - Calculus Proof

  1. #1
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    Calculus Proof

    I am having trouble using calculus to prove the following

    Show that: 0<a<b when 1/b < (ln b - ln a)/b-a < 1/a

    for example i know that 0<5<8 1/8 < (ln 8 - ln 5)/8-5 < 1/5 but i have no idea how to prove this

    please help?
    Last edited by greentea101; November 8th 2009 at 03:14 PM.
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  2. #2
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    Quote Originally Posted by greentea101 View Post
    I am having trouble using calculus to prove the following

    Show that: 0<a<b when 1/b < (ln b - ln a)/b-a < 1/a

    for example i know that 0<5<8 1/8 < (ln 8 - ln 5)/8-5 < 1/5 but i have no idea how to prove this

    please help?

    The wording is terrible: apparently you want to show that if 0<a<b , then \frac{1}{b}<\frac{\ln b-\ln a}{b-a}<\frac{1}{a}...**sigh**

    Ok, use Rolle's theorem for the derivable function f(x)=\ln x

    Tonio
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  3. #3
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    see if this helps

    if

    (1/b) < [(ln b - ln a)/(b-a)] < (1/a)

    then

    (1/b) < (1/a) -- by simply ignoring the middle expression of the inequality

    thus,

    b > a

    * now we must show that 'a' is greater than 0.

    if

    b > a

    then

    ln b > ln a

    thus,

    [(ln b - ln a)/(b-a)] > 0

    if

    0 < [(ln b - ln a)/(b-a)] < (1/a)

    then

    (1/a) > 0

    thus,

    a > 0

    and therefore,

    0 < a < b
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  4. #4
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    Quote Originally Posted by DanielJackson View Post
    if

    (1/b) < [(ln b - ln a)/(b-a)] < (1/a)

    then

    (1/b) < (1/a) -- by simply ignoring the middle expression of the inequality

    thus,

    b > a

    What if b<0 and a>0 ? Then it would be come a>b


    * now we must show that 'a' is greater than 0.

    if

    b > a

    then

    ln b > ln a

    If a<0 then ln a is not defined, so here you have assumed that 0<a<b which is
    the thing you are trying to prove.


    thus,

    [(ln b - ln a)/(b-a)] > 0

    if

    0 < [(ln b - ln a)/(b-a)] < (1/a)

    then

    (1/a) > 0

    thus,

    a > 0

    and therefore,

    0 < a < b
    This is not this simple.
    It looks like you could use the mean value theorem....but I dont know.
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