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Math Help - Algebra

  1. #1
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    Algebra

    X=Y e(to the power of) p x t

    Y and p are constant

    determine approx values of y and p

    when
    X = 75 t = 2.25
    X = 50 t = 1.25

    cheers for your help
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  2. #2
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    Quote Originally Posted by mathsisfun
    X=Y e(to the power of) p x t

    Y and p are constant

    determine approx values of y and p

    when
    X = 75 t = 2.25
    X = 50 t = 1.25

    cheers for your help
    There is a dictum in math that if the number of variables exceed the number of equations then you cannot find unique solutions. Over here you have two variable you need to solve for p and y. And only one equation.
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  3. #3
    Member
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    Wethersfield, CT
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    92
    Hi:

    From the given information, we can substitute real values for x and t, giving the equations 75 = ye^2.25p and 50 = ye^1.25p. Dividing the first equation by the second eliminates the constant y, and we have 75/e^2.25p = 50/e^1.25p. Cross multiply and solve for p. Finally, substitute the value of p, just determined, into either of the two equations derived in the step #1, and solve for y. I leave the algebra to you. Hint: The product of p and y is between 12 and 13.

    Regards,

    Rich B.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by mathsisfun
    X=Y e(to the power of) p x t

    Y and p are constant

    determine approx values of y and p

    when
    X = 75 t = 2.25
    X = 50 t = 1.25

    cheers for your help
    When x=75 and t=2.25,
    75 = y[e^(p*2.25)
    75/y = e^(2.25p)
    Take the natural logs of both sides,
    ln(75/y) = (2.25p)ln(e)
    ln(75) -ln(y) = 2.25p -----------------(1)

    When x=50 and t=1.25,
    50 = y[e^(p*1.25)
    50/y = e^(1.25p)
    Take the natural logs of both sides,
    ln(50) -ln(y) = 1.25p -----------------(2)

    Let's solve for p first,
    (1) minus (2),
    ln(75) -ln(50) = 2.25p -1.25p
    ln(75/50) = 1.00p
    p = ln(3/2) = ln(1.5) ----------------------***
    Substitute that into, say, (1),
    ln(75) -ln(y) = 2.25ln(1.5)
    ln(75) -ln(1.5^(2.25)) = ln(y)
    ln[75/(1.5^(2.25))] = ln(y)
    y = 75/(1.5^(2.25)) = 30.12 -------***

    Check those against (2),
    ln(50) -ln(30.12) =? 1.25(ln(1.5))
    3.912 -3.405 =? 0.507
    0.507 =? 0.507
    Yes, so, OK.

    Therefore, y=30.12 and p=ln(1.5). -----------answer.

    ---------------------
    Reminders:

    ----log(a/b) = log(a) -log(b).
    ----if log(a) = log(b), then a=b.
    ----c*log(a) = log(a^c).
    ----if ln(a) = b, then a = e^b, that is why ln(e) = 1, or log(10) = 1.
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  5. #5
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    Mar 2006
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    Thankyou very much for your help.
    I understand now, thanks for the reminders too, this is where I was going wrong so they are very useful.

    Cheers
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