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Logarithms disinfectant spray problem

  1. Dec 8, 2007 #1
    Question:
    A new disinfectant spray is expected to kill 50% of the known germs in a room, but for health reasons it can only be used once a day. Between spraying, the germs increase by 25%. How many consecutive days of spraying are required to reduce the germs in the room to 10% of the original amount?


    Relevant equations:
    A=Ao(1+i)^n


    Attempt:
    0.10Po=Po(0.50)^d + 0.25Po(0.50)^d
    0.10=0.50^d + 0.25(0.50)^d
    0.10=1.25(0.50)^d
    0.08=0.50^d
    log0.08=log0.50^d
    log0.08=dlog0.50
    d=[tex]\frac{log0.08}{log0.50}[/tex]
    d=4

    Can someone please tell me what I'm doing wrong? The answer is supposed to be 5 days.
     
  2. jcsd
  3. Dec 8, 2007 #2
    what do the variables of your relevant equation pertain to?
     
  4. Dec 8, 2007 #3
    A is the final amount. Ao is the initial amount. i is the increase or decrease. and n is the number of periods.
     
  5. Dec 8, 2007 #4
    i'm unable to follow your problem but when we solve for n, which i assume is the number of days

    [tex]A=A_{0}(1+i)^{n}[/tex]

    divide by A initial and then take the log of both sides and then solve for n
     
  6. Dec 8, 2007 #5
    yea, i tried doing that. but the answer didn't come out right.
     
  7. Dec 8, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If A is the number of germs then after spraying A->0.5*A. After waiting a day that number increases by 25%. So that's multiplication by 1.25. Put the two together and from one day to the next A->0.5*A*1.25=0.625*A. So that's 0.1=(0.625)^n.
     
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