Answer:
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.
Step-by-step 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 :
[tex]A=10*(0.82)^{t}[/tex]
a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:
[tex]A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\[/tex]
A=10
The initial amount was 10 mg.
b. Considering that an exponential growth is determined by:
[tex]A=A0*(1-r)^{t}[/tex], 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:
[tex]A=10*(0.82)^{6}[/tex]
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:
[tex]1=10*(0.82)^{t}[/tex]
Solving:
[tex]0.1=(0.82)^{t}[/tex]
㏒ 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.
