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Math Help - Exponential algebra

  1. #1
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    Exponential algebra

    Howdy folks. First time poster here

    Have a bit of a problem and can't seem to figure the correct way to solve it

    I'm looking for Ka

    0.93Ka - 1.969 = -e^(-2Ka)

    Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.
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  2. #2
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    Smile

    just for curiosity,is this problem has anything to do with chemistry ?
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  3. #3
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    Quote Originally Posted by Raoh View Post
    just for curiosity,is this problem has anything to do with chemistry ?
    Sort of. It's to do with the calculation of the reaeration factor for a lake.

    Any ideas?
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  4. #4
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    Smile

    I thought it has something to do with acids,'cause there's this interesting relation PKa=-\log Ka.
    (Acidity constant).
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  5. #5
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    Hello FreemasonFurbie

    Welcome to Math Help Forum!
    Quote Originally Posted by FreemasonFurbie View Post
    Howdy folks. First time poster here

    Have a bit of a problem and can't seem to figure the correct way to solve it

    I'm looking for Ka

    0.93Ka - 1.969 = -e^(-2Ka)

    Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.
    You won't be able to find an algebraic solution. Your best bet is a numerical method. With a spreadsheet, I've just found a couple of solutions: one negative and one positive. To 4 d.p. they are:
    K_a=-0.4315, \;+2.1011
    I hope that helps.

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello FreemasonFurbie

    Welcome to Math Help Forum!
    You won't be able to find an algebraic solution. Your best bet is a numerical method. With a spreadsheet, I've just found a couple of solutions: one negative and one positive. To 4 d.p. they are:
    K_a=-0.4315, \;+2.1011
    I hope that helps.

    Grandad
    Thanks! Would you mind explaining how you did it in Excel? I'd be interested in seeing how you did it.

    I wasn't aware that it had such a function. I was going to use Matlab, but alas, I forgot to install it on my computer after my last restore.
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  7. #7
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    Hello FreemasonFurbie
    Quote Originally Posted by FreemasonFurbie View Post
    Thanks! Would you mind explaining how you did it in Excel? I'd be interested in seeing how you did it.

    I wasn't aware that it had such a function. I was going to use Matlab, but alas, I forgot to install it on my computer after my last restore.
    Nothing very subtle - trial and error, really. I just set up the function 0.93x+exp(-2x) and tried some likely values around x=0. Then refined them when I got a general feel for the behaviour of the function.

    Without a spreadsheet I should use Newton's Method - but to be honest it's loads quicker just to stick some numbers into Excel! (You'd never think I was a real mathematician, would you?)

    Spreadsheet attached (for what it's worth!)

    Grandad
    Attached Files Attached Files
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  8. #8
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    You chancer!!!

    haha.....

    Thanks Grandad. The help is much appreciated!
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by FreemasonFurbie View Post
    Howdy folks. First time poster here

    Have a bit of a problem and can't seem to figure the correct way to solve it

    I'm looking for Ka

    0.93Ka - 1.969 = -e^(-2Ka)

    Any ideas in how to solve it? I was planning on taking LN of both sides, but that still leaves me with LN of Ka on one side with Ka on the other side of the equation, which is practically the same.
    Solve numerically the solution in (1,5) is: Ka\approx 2.101

    Can also be solved in terms of Lambert's W function:

    Ka=\frac{1}{2}\;W\left[-\frac{2}{a}\; e^{-2b/a}\right]+\frac{b}{a}

    CB
    Last edited by CaptainBlack; December 3rd 2009 at 12:08 AM.
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