At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by A = 12(0.9). (a) What was the initial amount taken? 12 mg (b) What percent of the drug leaves the body each hour? Decimal form: % (c) Find the time until only 1 mg of the drug remains in the body. Give your answer in exact form and decimal form. Exact form: hours hours (nearest tenth)
At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by A = 12(0.9). (a) What was the initial amount taken? 12 mg (b) What percent of the drug leaves the body each hour? Decimal form: % (c) Find the time until only 1 mg of the drug remains in the body. Give your answer in exact form and decimal form. Exact form: hours hours (nearest tenth)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Drug Decay Model: Hydrocodone Bitartrate
At time \( t \) hours after taking the cough suppressant hydrocodone bitartrate, the amount, \( A \) in mg, remaining in the body is given by the equation:
\[ A = 12(0.9)^t \]
#### Question (a)
**What was the initial amount taken?**
*Answer:*
[ ✔ ] \( \boxed{12} \) mg
#### Question (b)
**What percent of the drug leaves the body each hour?**
*Answer:*
[ ✘ ] (This answer was not provided)
#### Question (c)
**Find the time until only 1 mg of the drug remains in the body. Give your answer in exact form and decimal form.**
*Answer:*
- **Exact form:** [ ] hours
- **Decimal form:** [ ] hours (nearest tenth)
---
#### Explanation of the Equation and Process:
1. **Recognizing Initial Amount:**
- From the equation \( A = 12(0.9)^t \), the initial amount \( A \) when \( t=0 \) is \( 12 \) mg.
2. **Percent Decay per Hour:**
- The base of the exponential, \( 0.9 \), indicates that 90% of the drug remains each hour.
- Therefore, \( 100\% - 90\% = 10\% \) of the drug leaves the body every hour.
3. **Solving for Time when 1 mg Remains:**
- To find the time \( t \) when \( A = 1 \) mg:
\[
1 = 12(0.9)^t
\]
Divide both sides by 12:
\[
(0.0833) = (0.9)^t
\]
Taking the natural logarithm of both sides:
\[
\ln(0.0833) = t \ln(0.9)
\]
Solving for \( t \):
\[
t = \frac{\ln(0.0833)}{\ln(0.9)}
\]
The exact and decimal form answers can be calculated using a scientific calculator or natural logarithm tables.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedc716be-e066-4234-bd1a-1a86837575c3%2Fafd331c0-f88c-49ad-bd8f-3a0095cbffbc%2Fdeimzww_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
### Drug Decay Model: Hydrocodone Bitartrate
At time \( t \) hours after taking the cough suppressant hydrocodone bitartrate, the amount, \( A \) in mg, remaining in the body is given by the equation:
\[ A = 12(0.9)^t \]
#### Question (a)
**What was the initial amount taken?**
*Answer:*
[ ✔ ] \( \boxed{12} \) mg
#### Question (b)
**What percent of the drug leaves the body each hour?**
*Answer:*
[ ✘ ] (This answer was not provided)
#### Question (c)
**Find the time until only 1 mg of the drug remains in the body. Give your answer in exact form and decimal form.**
*Answer:*
- **Exact form:** [ ] hours
- **Decimal form:** [ ] hours (nearest tenth)
---
#### Explanation of the Equation and Process:
1. **Recognizing Initial Amount:**
- From the equation \( A = 12(0.9)^t \), the initial amount \( A \) when \( t=0 \) is \( 12 \) mg.
2. **Percent Decay per Hour:**
- The base of the exponential, \( 0.9 \), indicates that 90% of the drug remains each hour.
- Therefore, \( 100\% - 90\% = 10\% \) of the drug leaves the body every hour.
3. **Solving for Time when 1 mg Remains:**
- To find the time \( t \) when \( A = 1 \) mg:
\[
1 = 12(0.9)^t
\]
Divide both sides by 12:
\[
(0.0833) = (0.9)^t
\]
Taking the natural logarithm of both sides:
\[
\ln(0.0833) = t \ln(0.9)
\]
Solving for \( t \):
\[
t = \frac{\ln(0.0833)}{\ln(0.9)}
\]
The exact and decimal form answers can be calculated using a scientific calculator or natural logarithm tables.
---
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