2000(1 – 0.75) = 2000 (.25)
= 500 survivors after the dose of prescription a is taken
Then,
500(1.40)3 = 1374 bacteria after the 3 hours period at the instant before the second dose is taken.
1.) For antibiotic A, determine the bacterial population after nine hours (before the fourth dose is taken)
2.) Another antibiotic, Antibiotic B, is taken every 6 hours and has an effectiveness of 90%. Use a process similar to (1) to find the equation N(t). Then use N(t) to determine the bacterial population after 3 days (72 hours) in a patient who has an initial bacterial population of 2000 and who is being treated with Antibiotic B.
3.) A third antibiotic, antibiotic C, has an effectiveness of 99.5%. Explain why antibiotic C cannot cure a person if it is administered every 24 hours
To solve these problems, we need to use the concept of exponential decay or growth. The formula for exponential decay is given by:
N(t) = N0 * e^(-kt)
Where:
N(t) = Final population at time t
N0 = Initial population (at time t=0)
k = Decay constant (or growth constant if k is positive)
t = Time
Now let's solve each problem step by step:
1.) For antibiotic A, determine the bacterial population after nine hours (before the fourth dose is taken)
Given:
Initial population (N0) = 1374
Time (t) = 9 hours
k = Unknown
We can rearrange the formula to solve for k:
N(t) = N0 * e^(-kt)
1374 = 1374 * e^(-k * 9)
e^(-9k) = 1
-9k = ln(1)
k = 0
Since k = 0, it means there is no decay or growth. Therefore, the bacterial population remains the same after nine hours. So the answer is 1374.
2.) For antibiotic B, let's find the equation N(t). We know that it's taken every 6 hours with an effectiveness of 90%. This means the decay constant (k) is given by:
k = -ln(0.90) ≈ 0.1054
Now, let's find the bacterial population after 3 days (72 hours) with an initial population of 2000:
Given:
Initial population (N0) = 2000
Time (t) = 72 hours
k = 0.1054
Plugging these values into the formula:
N(t) = N0 * e^(-kt)
N(72) = 2000 * e^(-0.1054 * 72)
Using a calculator, we find:
N(72) ≈ 2,000 * 0.000499 ≈ 0.998 ≈ 1
So, the bacterial population after 3 days is approximately 1.
3.) For antibiotic C with an effectiveness of 99.5% and administered every 24 hours, we can follow a similar process as before.
Given:
Effectiveness = 99.5% = 0.995
Time (t) = 24 hours
k = -ln(0.995)
To solve for k, we have:
k ≈ 0.00501
Now, let's assume the initial population (N0) is 2000. Plugging these values into the formula:
N(t) = N0 * e^(-kt)
N(24) = 2000 * e^(-0.00501 * 24)
Using a calculator, we find:
N(24) ≈ 2000 * 0.818 ≈ 1636
After 24 hours, antibiotic C reduces the population to approximately 1636 bacteria. However, it does not cure the person since it doesn't eliminate all the bacteria. The remaining 1636 bacteria can still multiply and cause infections. To achieve a complete cure, a higher effectiveness or more frequent dosing might be required.