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Math Help - Annoying formula. (Terrible description, sorry)

  1. #1
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    Annoying formula. (Terrible description, sorry)

    Hi all, normally pretty good with day to day math, but bit off more than I can chew on this one, not entirely sure where I should post the question as it's not for school, so please accept apologies if it's in the wrong place.

    I'm trying to reverse (if that's an applicable description) this equation so that I can find 'x' if I have 'a', 'y' & 'z'. Unfortunately I'm more of an idiot triangle kinda guy for that sorta thing, and it doesn't seem to be an option on this one.

    a=495/(1.0324-.19077(log(x-y))+.15456(log(z)))-450


    After playing around for half an hour, I'm getting nowhere, and my brain hurts, so I figured I'd bow to your superior intellects, and join the forum to beg for help.

    No hurry, as it's just one of those random bright ideas I got, which is of little practical use. I've figured out the answer eventually through cheating with entering random numbers into excel for 'x' until I got an answer close enough to suit my curiosity. But it's annoying me that I can't figure it out the formula, which is now a bigger problem than getting the answer, as I don't like not being able to do things.
    Hopefully you can sort me out, thanks in advance for any help.
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  2. #2
    Senior Member pacman's Avatar
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    ambrosius, i will start, you finish it . . . for x

    a = (495)/(1.0324 - 0.19077log(x - y) + 0.15456(log z)) - 450

    (a - 450) = 495/(1.0324 - 0.19077log(x - y) + 0.15456(log z))

    (1.0324 - 0.19077log(x - y) + 0.15456(log z)) = 495/(a - 450)

    - 0.19077log(x - y) = 495/(a - 450) - 1.0324 - 0.15456(log z)

    log(x - y) = [495/(a - 450) - 1.0324 - 0.15456(log z)]/(-0.19077),

    you can finish it off . . .
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  3. #3
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    Lol, I really shoulda concentrated more in High School. Finally got it, in theory, seems to give the expected answer within a suitable margin for error in the decimals.

    x=10^((1.0324-((495/(a+450))-(0.15456*LOG(z*100))))/0.19077))+y)

    Thanks for the help, and remind me never to try something like that again.
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