(10 points) The amount of doxycyclin (an antibiotic) decays rapidly in the blood.
If a patient takes 200 mg of doxycyclin, it can be measured that after 5 hours she will have only 50 mg left in her blood. Assuming that the amount of doxycyclin in her blood is f(x) = Abx, where x is the number of hours elapsed,
(a) find A and b.
(b) How much does she have left after 6 hours?
(c) What percentage is removed every hour? (d) When will be 30 mg left?

We are given that the amount of doxycycline in the blood after \( x \) hours can be modeled by the function \( f(x) = A b^x \), where \( A \) is the initial amount and \( b \) is the decay factor per hour.

### Part (a) Find \( A \) and \( b \):

- The initial dose of doxycycline is 200 mg. Thus, \( A = 200 \), as the amount of doxycycline when \( x = 0 \) is 200 mg.

So, the function becomes:
\[
f(x) = 200 b^x
\]

- After 5 hours, the amount of doxycycline left is 50 mg. Thus, \( f(5) = 50 \).

Substitute \( f(5) = 50 \) and solve for \( b \):
\[
50 = 200 b^5
\]
\[
\frac{50}{200} = b^5
\]
\[
0.25 = b^5
\]
Now, solve for \( b \) by taking the fifth root:
\[
b = (0.25)^{\frac{1}{5}}
\]
\[
b = 0.72478
\]

Thus, \( A = 200 \) and \( b \approx 0.7248 \).

### Part (b) How much is left after 6 hours?

We use the function \( f(x) = 200 (0.7248)^x \) to find the amount of doxycycline after 6 hours:

\[
f(6) = 200 (0.7248)^6
\]
Let me calculate that.

\[
f(6) \approx 200 \times 0.7248^6
\]

Let me evaluate this.After 6 hours, the patient will have approximately 29 mg of doxycycline left in her blood.

### Part (c) What percentage is removed every hour?

The decay factor per hour is \( b \approx 0.7248 \). This means that after each hour, 72.48% of the remaining doxycycline is left. The percentage removed each hour is:
\[
\text{Percentage removed} = (1 - b) \times 100
\]
\[
\text{Percentage removed} = (1 - 0.7248) \times 100 = 27.52\%
\]
Thus, about 27.52% of the doxycycline is removed from the blood every hour.

### Part (d) When will there be 30 mg left?

We want to find \( x \) such that \( f(x) = 30 \):
\[
30 = 200 \times (0.7248)^x
\]
Solve for \( x \):
\[
\frac{30}{200} = (0.7248)^x
\]
\[
0.15 = 0.7248^x
\]
Take the natural logarithm of both sides:
\[
\ln(0.15) = x \ln(0.7248)
\]
Now solve for \( x \):
\[
x = \frac{\ln(0.15)}{\ln(0.7248)}
\]

Let me calculate this value for \( x \).There will be approximately 30 mg of doxycycline left in the patient's blood after about 5.89 hours (close to 5 hours and 53 minutes).

### Summary:
(a) \( A = 200 \), \( b \approx 0.7248 \)  
(b) After 6 hours, approximately 29 mg is left.  
(c) About 27.52% of the doxycycline is removed every hour.  
(d) There will be 30 mg left after approximately 5.89 hours.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the decay function change if the initial dose were 150 mg instead of 200 mg?
2. What is the half-life of doxycycline in this model?
3. How much doxycycline would remain after 10 hours?
4. How would the percentage of doxycycline removed change if the decay factor was lower?
5. Can we express the decay model as a continuous exponential decay?

**Tip:** Exponential decay models are widely used in pharmacology to represent how drugs are metabolized and cleared from the bloodstream.

The amount of doxycyclin (an antibiotic) decays rapidly in the blood. If a patient takes 200 mg of doxycyclin, it can be measured that after 5 hours she will have only 50 mg left in her blood. Assuming that the amount of doxycyclin in her blood is f(x) = Ab^x, where x is the number of hours elapsed, (a) find A and b. (b) How much does she have left after 6 hours? (c) What percentage is removed every hour? (d) When will be 30 mg left?

Explore how the amount of doxycyclin in the blood decays over time with an exponential function. Learn how to calculate the initial dose, decay factor, and the percentage of the drug removed every hour, with detailed steps. Suitable for high school students in Grades 10-12.

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