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Math Help - 80% to 95% - How much lysol will you need?

  1. #1
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    80% to 95% - How much lysol will you need?

    Spraying 18.0 mg of lysol will kill 80% of bacteria on the table. How much lysol must be sprayed to kill 95% of these bacteria on the table. (it will have an exponential relationsip)
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  2. #2
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    If you spray 18mg, there are 20% of the bacteria remaining. If you spray another 18mg, it will kill all but 20% of the remaining bacteria, so there will be 20% * 20% = 4% remaining. Let r be the portion of bacteria remaining and x the amount of lysol sprayed. Then

    r = (20%)^(x/18mg)

    So we want to find the x that makes r=5%. As a check, since 36mg kills just a little too much, your answer should be slightly less than 36mg.

    Post again in this thread if you're still having trouble.

    - Hollywood
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    Quote Originally Posted by hollywood View Post
    If you spray 18mg, there are 20% of the bacteria remaining. If you spray another 18mg, it will kill all but 20% of the remaining bacteria, so there will be 20% * 20% = 4% remaining. Let r be the portion of bacteria remaining and x the amount of lysol sprayed. Then

    r = (20%)^(x/18mg)

    So we want to find the x that makes r=5%. As a check, since 36mg kills just a little too much, your answer should be slightly less than 36mg.

    Post again in this thread if you're still having trouble.

    - Hollywood
    Thanks. This is more clear.

    How would we get the answer using differentials (like dP/dt)? and the exponential e?
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  4. #4
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    Ok. Let r be the fraction of bacteria remaining, and let x be the amount of lysol sprayed. If you spray a very small amount, it will kill an amount of bacteria proportional to the fraction of bacteria remaining:

    \frac{dr}{dx}=-kr, where k is the constant of proportionality. Note the negative sign - as x increases, r decreases.

    The solution is r=Ce^{-kx}. Since not spraying any lysol results in no bacteria killed, r=1 when x=0, so C=1.

    We also know that r=0.2 when x=18mg, so

    0.2=e^{-k(18mg)}

    k=-\frac{\ln{0.2}}{18mg}

    So we want to know what x gives r=0.05.

    0.05=e^{-kx}

    x=-\frac{\ln{0.05}}{k}=\frac{\ln{0.05}}{\ln{0.2}}(18m  g) or approximately 33.5mg.

    It's essentially the same calculation as before.

    - Hollywood
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