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Math Help - Helpp exponent & logarithmic problems!!

  1. #1
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    Helpp exponent & logarithmic problems!!

    sirs Sorry for this long questions in my post because its already our finals and i cant seem to answer our reviewer with so many undiscussed LOg problems!!! i wont do this again....
    U can answer as many...

    Exponential growth / decay
    2.) In year 2000, it was expected that for the succeeding 20 years the population of a particular town would be y people t years from 2000, where y = C10 ^ (kt) and C and k are constants. If the actual population in 2000 was 10,000 and in 2005 it was 40,000 what is the expected population in 2015?

    3.) Suppose the amount of oil pumped from one of the oils wells in jeddah decreases at the continuous rate of 10% per year. When will the well's output fall to one-fifth of its present value? use the typical equation growth decay.

    Heat transfer: Newton's Law of Cooling
    4) Soup Left in a tin cup cools to the temperature of the surrounding air. a hot silver ingot immersed in water cools to the temperature of the surrounding water. In situations like these, the temperature of a hot object changes w/ time according to the newton's law of cooling:
    T - Ts = ( To - Ts) e ^ -kt
    where T is the temperature of the hot object at time t, Ts is the surrounding temperature that is assumed constant, To is the value of T at time zero and k is called the heat transfer coefficient, also a constant. The equation is also applicable to the case of a warming object.

    Now suppose a hard - boiled egg at 98 degrees Celsius water to cool. After 5 minutes, the egg's temperature is found to be 38 degrees Celsius. Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach 20 degrees celsius?

    Electric Circuits
    6.) A simple electric circuit containing no condensers, a resistance of R ohms, and an inductance of L henries has the electromotive force cut off when the current is Io amperes. The current dies down so that at t seconds the current is i amperes, and i = Io e^ -(R/L)t. Use natural logarithms to solve this equation for t in terms of i and the constants R, L and Io.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    2) In year 2000, it was expected that for the succeeding 20 years the population of a particular town would be y people t years from 2000, where y = C10 ^ (kt) and C and k are constants. If the actual population in 2000 was 10,000 and in 2005 it was 40,000 what is the expected population in 2015?
    You have
    10000 = C*10^(0*k)
    40000 = C*10^(5k)

    From the first equation:
    10000 = C*10^0 = C*1 = C
    so C = 10000

    The second equation becomes:
    40000 = 10000*10^(5k)

    4 = 10^(5k)

    log 4 = 5k <-- log base 10

    or k = (1/5)*log 4 (is approximately) 0.1204

    So in 2015:
    y = 10000*10^(0.1204*15) = 640000

    (If you wish to do this exactly:
    y = 10000*10^[(1/5)*(log 4)*15]
    y = 10000*10^[3*(log 4)]
    y = 10000*{10^(log4)}^3
    y = 10000*{4}^3 = 10000*64
    y = 640000).

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    3.) Suppose the amount of oil pumped from one of the oils wells in jeddah decreases at the continuous rate of 10% per year. When will the well's output fall to one-fifth of its present value? use the typical equation growth decay.
    A 10% decline per year means that if we call y the amount of oil pumped from one oil well in a year and t is the time in years:
    y = C*10^(-t)

    When the output is 1/5 of the present (t = 0) value then y = (1/5)*C.

    (1/5)*C = C*10^(-t)

    1/5 = 10^(-t)

    This is solved the same way as problem 2. (I got t = -log (1/5) = log 5.)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Heat transfer: Newton's Law of Cooling
    4) Soup Left in a tin cup cools to the temperature of the surrounding air. a hot silver ingot immersed in water cools to the temperature of the surrounding water. In situations like these, the temperature of a hot object changes w/ time according to the newton's law of cooling:
    T - Ts = ( To - Ts) e ^ -kt
    where T is the temperature of the hot object at time t, Ts is the surrounding temperature that is assumed constant, To is the value of T at time zero and k is called the heat transfer coefficient, also a constant. The equation is also applicable to the case of a warming object.

    Now suppose a hard - boiled egg at 98 degrees Celsius (is placed in) water to cool. After 5 minutes, the egg's temperature is found to be 38 degrees Celsius. Assuming that the water has not warmed appreciably, how much longer will it take the egg to reach 20 degrees celsius?
    The problem is we don't know the temperature of the water bath, so we can't finish the problem. I presume there was a typo in the problem and that you have this value. (If you don't, the problem can't be solved numerically.) But I'll set it up for you.
    We first need to find k.
    t = 5 min
    To = 98 C, T = 38 C
    38 - Ts = (98 - Ts)*e^(-k*5)
    (38 - Ts)/(98 - Ts) = e^(-5k)

    You can solve this for k. (Hint use log to the base e, or "ln" instead of log to the base 10.)

    For the egg to reach 20 C we have
    To = 38 C, T = 20 C (We need to find "how much longer," so I'm resetting the clock)

    20 - Ts = (38 - Ts)*e^(-k*t) <-- Using the k value above.

    Which you can solve now for t.

    If you aren't given a value for Ts, then the solution is
    t = 5*{ln[(38-Ts)/(20-Ts)]}/{ln[(98-Ts)/(38-Ts)]}

    or you can write this as:
    t = 5*[ln(38-Ts) - ln(20-Ts)]/[ln(98-Ts) - ln(38-Ts)]

    -Dan
    Last edited by topsquark; October 2nd 2006 at 05:59 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Electric Circuits
    6.) A simple electric circuit containing no condensers, a resistance of R ohms, and an inductance of L henries has the electromotive force cut off when the current is Io amperes. The current dies down so that at t seconds the current is i amperes, and i = Io e^ -(R/L)t. Use natural logarithms to solve this equation for t in terms of i and the constants R, L and Io.
    You should be able to solve this on your own given the techniques above. I get that
    t = (L/R)*ln(Io/i)
    If you have problems with this, just let me know.

    -Dan
    Last edited by topsquark; October 2nd 2006 at 05:52 AM. Reason: Flipped the coefficient
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  6. #6
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    Thanks Dan!!!!!!!!!!
    Ur really a genius!!!
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  7. #7
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    Quote Originally Posted by topsquark View Post
    The problem is we don't know the temperature of the water bath, so we can't finish the problem. I presume there was a typo in the problem and that you have this value. (If you don't, the problem can't be solved numerically.) But I'll set it up for you.
    We first need to find k.
    t = 5 min
    To = 98 C, T = 38 C
    38 - Ts = (98 - Ts)*e^(-k*5)
    (38 - Ts)/(98 - Ts) = e^(-5k)

    You can solve this for k. (Hint use log to the base e, or "ln" instead of log to the base 10.)

    For the egg to reach 20 C we have
    To = 38 C, T = 20 C (We need to find "how much longer," so I'm resetting the clock)

    20 - Ts = (38 - Ts)*e^(-k*t) <-- Using the k value above.

    Which you can solve now for t.

    If you aren't given a value for Ts, then the solution is
    t = 5*{ln[(38-Ts)/(20-Ts)]}/{ln[(98-Ts)/(38-Ts)]}

    or you can write this as:
    t = 5*[ln(38-Ts) - ln(20-Ts)]/[ln(98-Ts) - ln(38-Ts)]

    -Dan
    sorry
    ... at 98 degrees is put in a sink of 18 degrees C water to cool...
    Last edited by ^_^Engineer_Adam^_^; October 2nd 2006 at 09:16 AM.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    The problem is we don't know the temperature of the water bath, so we can't finish the problem. I presume there was a typo in the problem and that you have this value. (If you don't, the problem can't be solved numerically.) But I'll set it up for you.
    We first need to find k.
    t = 5 min
    To = 98 C, T = 38 C
    38 - Ts = (98 - Ts)*e^(-k*5)
    (38 - Ts)/(98 - Ts) = e^(-5k)

    You can solve this for k. (Hint use log to the base e, or "ln" instead of log to the base 10.)

    For the egg to reach 20 C we have
    To = 38 C, T = 20 C (We need to find "how much longer," so I'm resetting the clock)

    20 - Ts = (38 - Ts)*e^(-k*t) <-- Using the k value above.

    Which you can solve now for t.

    If you aren't given a value for Ts, then the solution is
    t = 5*{ln[(38-Ts)/(20-Ts)]}/{ln[(98-Ts)/(38-Ts)]}

    or you can write this as:
    t = 5*[ln(38-Ts) - ln(20-Ts)]/[ln(98-Ts) - ln(38-Ts)]

    -Dan
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    sorry
    ... at 98 degrees is put in a sink of 18 degrees C water to cool...
    Okay, so the equation for k becomes:

    (38 - 18)/(98 - 18) = e^(-5k)

    1/4 = e^(-5k) which give me k = (1/5)*ln 4 which is approx. = 0.2773/min.

    Then the t equation becomes:
    20 - 18 = (38 - 18)*e^(-0.2773*t)
    which gives me t = 8.305 min.

    (Note: I used the exact expression for k to get this t value. If you used an approximation you'll get a slightly different value. I got that t = 5*(ln 10)/(ln 4) for the exact expression.)

    -Dan
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