Question

At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by

A=10(0.82)^t

Required:
a. What was the initial amount taken?
b. What percent of the drug leaves the body each hour?
c. How much of the drug is left in the body 6 hours after the dose is administered?
d. How long is it until only 1 mg of the drug remains in the body?

Answer 1

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time until only 1 mg of the drug remains in the body is 11.6 hours.

Explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

`A=10*(0.82)^(t)`

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

`A=10*(0.82)^(t)=10*(0.82)^(0)=10*1\\`

A=10

The initial amount was 10 mg.

b. Considering that an exponential growth is determined by:

`A=A0*(1-r)^(t)`, where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

`A=10*(0.82)^(6)`

Solving:

A= 3.04 mg

The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

`1=10*(0.82)^(t)`

Solving:

`0.1=(0.82)^(t)`

㏒ 0.1= t*㏒ 0.82

㏒ 0.1 ÷ ㏒ 0.82= t

11.6 hours= t

The time until only 1 mg of the drug remains in the body is 11.6 hours.