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Math Help - Hi Guys! algebra help needed for engineering studies!

  1. #1
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    Hi Guys! algebra help needed for engineering studies!

    Hi there Im andy from Liverpool
    im currently studying for an engineering degree, but firstly have to complete a foundation course in maths. Its going well so far - until the dreaded algebra questions came along! Ive understood some of it so far but the assesment questions have tied me in knots and i feel completely stuck. These are the questions that are giving me particular difficulty, any help would be much appreciated.
    Solve: 16 = 24(1-e-t) (It shgould read -t over 2 but cant figure out to write correctly on here!)
    2

    The next question is

    A= V (Q- mV2) Where Q = 50.28, m = 17, V = 5 and g = 9.81
    100 g

    Many thanks
    Andy
    Last edited by andym1512; February 18th 2013 at 02:46 PM.
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  2. #2
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    Re: Hi Guys! algebra help needed for engineering studies!

    Sorry second equation isnt very clear should be V divided by 100 mV squared divided by g
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: Hi Guys! algebra help needed for engineering studies!

     16 = 24(1-e^{-\frac{t}{2}})
    divide both sides by 24 to get
     \frac{2}{3} = 1 -e^{-\frac{t}{2}}
    add - 1 to both sides
     \frac{-1}{3} = -e^{-\frac{t}{2}}
    multiply both sides by -1
     \frac{1}{3} = e^{-\frac{t}{2}})
    take natural log of both sides
    ln(\frac{1}{3}) = -\frac{t}{2}
    use log rules a*ln(x) = ln(x^a)
    to get 2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})
    so -t = ln(\frac{1}{9})
    t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)
    t = ln(9)
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  4. #4
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    Re: Hi Guys! algebra help needed for engineering studies!

    thanks very much for reply, really helped!
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  5. #5
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    Re: Hi Guys! algebra help needed for engineering studies!

    Quote Originally Posted by jakncoke View Post
     16 = 24(1-e^{-\frac{t}{2}})
    divide both sides by 24 to get
     \frac{2}{3} = 1 -e^{-\frac{t}{2}}
    add - 1 to both sides
     \frac{-1}{3} = -e^{-\frac{t}{2}}
    multiply both sides by -1
     \frac{1}{3} = e^{-\frac{t}{2}})
    take natural log of both sides
    ln(\frac{1}{3}) = -\frac{t}{2}
    use log rules a*ln(x) = ln(x^a)
    to get 2*ln(\frac{1}{3}) = ln((\frac{1}{3})^2) = ln(\frac{1}{9})
    so -t = ln(\frac{1}{9})
    t = - ln(\frac{1}{9})=ln((\frac{1}{9})^{-1}) = ln(9)
    t = ln(9)

    you are a genius..
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