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Math Help - Help with an equation.

  1. #1
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    Hi

    Im Have this equation and I have the result but I canīt get from a to b....
    Attached Thumbnails Attached Thumbnails Help with an equation.-first-half-equation.jpg   Help with an equation.-second-half-equastion.jpg  
    Last edited by mr fantastic; May 4th 2010 at 02:33 AM.
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  2. #2
    MHF Contributor
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    Hi

    There are several errors in your calculations

    At the end of the 1st page there is a + instead of a - inside the square root. So is it in the 2 first equations in the second page

    Then you made an error in the second equation of the second page in the simplification by \alpha

    The right equation is \beta e^{-\alpha r^V} = \frac{1+\sqrt{1+k_2 \alpha^2}}{k_1 \alpha}

    But I agree with you : the disappearance of \beta is strange
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    There are several errors in your calculations

    At the end of the 1st page there is a + instead of a - inside the square root. So is it in the 2 first equations in the second page

    Then you made an error in the second equation of the second page in the simplification by \alpha

    The right equation is \beta e^{-\alpha r^V} = \frac{1+\sqrt{1+k_2 \alpha^2}}{k_1 \alpha}

    But I agree with you : the disappearance of \beta is strange
    Thank you SOOOOOO MUUUUCH

    But what will you get r^v equal to??
    Will you devide the right side first with beta and then isolate r^v?

    Thank you for your help.
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  4. #4
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    Yes, once you have e^{-\alpha r^V}= \frac{1+\sqrt{1+ k_2\alpha^2}}{\beta k_1\alpha}, take the logarithm of both sides.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Yes, once you have e^{-\alpha r^V}= \frac{1+\sqrt{1+ k_2\alpha^2}}{\beta k_1\alpha}, take the logarithm of both sides.
    No You cant just do that....it is the whole fraction you have to divide with beta...
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  6. #6
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    Never mind the last message ....Have found out that you are right. Sorry and thanks....
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