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At time (t) hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by A = 14(0.8)^t.. What percent of the drug leaves the body each hour?. How much of the drug is left in the body 2 hours after the dose is administered? How long is it until only 1 mg of the drug remains in the body?

User Razor Jack
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2 Answers

1 vote

Final answer:

The percent of the drug that leaves the body each hour is approximately 20%. Two hours after the dose is administered, approximately 8.96 mg of the drug is left in the body. It takes approximately 24.8 hours until only 1 mg of the drug remains in the body.

Step-by-step explanation:

To find the percent of the drug that leaves the body each hour, we can calculate the difference in the amount of the drug remaining after one hour and the initial amount. The initial amount of the drug is given by A = 14(0.8)^0, which is 14 mg. After one hour, the amount remaining is given by A = 14(0.8)^1, which is approximately 11.2 mg. The percent that leaves the body each hour is the difference between the initial amount and the amount remaining after one hour, divided by the initial amount, multiplied by 100. So, the percent that leaves the body each hour is (14 - 11.2) / 14 * 100, which is approximately 20%.

To find how much of the drug is left in the body 2 hours after the dose is administered, we substitute t = 2 into the equation A = 14(0.8)^t. So, A = 14(0.8)^2 = 14(0.64) = 8.96 mg.

To find how long it is until only 1 mg of the drug remains in the body, we need to solve the equation 1 = 14(0.8)^t for t. Divide both sides of the equation by 14 to get 0.0714 = (0.8)^t. Take the logarithm with base 0.8 of both sides to get t = log0.8(0.0714). Using a calculator, t is approximately 24.8 hours.

User Mikolaj
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8.5k points
3 votes
1)we calculate the amount of the drug is left in the body 2 hours after the dose is administered.
t=2
A(t)=14(08)^t
A(t)=14(0.8)²=14(0.64)=8.96

Answer: after of two hours the amount of drug left in te body is 8.96 mg.

2)we calculate the time required to stay in the body 1 mg of drug.

A(t)=1
14(0.8)^t=1
ln [14(0.8)^t]=ln1
ln14+tln(0.8)=0
tln(0.8)=-ln 14
t=-ln 14 / ln (0.8)
t=11.8267...≈11.83

Answer: the time required is 11.83 hours (≈11 hours, 49 minutes, 48 seconds).

User Ethereal
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8.6k points
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