A. If the half-life of the drug is 7.3 hours, what fraction of the drug remains
ID: 2968822 • Letter: A
Question
A.If the half-life of the drug is 7.3 hours, what fraction of the drug remains in the patient after 24 hours?
fraction =
D.What is the long-term minimum amount of drug in the patient?
Long-term minimum = mg
Explanation / Answer
Concentration of the drug declines as C = Co / 2^(t / T), where T is the half-life.
A) C = Co / 2^(12 hr / 7.1 hr) --> C/Co = 1 / 2^(12/7.1)
B) If An is the concentration immediately after the nth dose, then
An = A1 + (An-1) / 2^(12/7.1)
where A1 is of course the 60 mg dose.
C) You haven't said what n1 is. If Pn is the concentration immediately before the nth dose,
Pn = Pn-1 / 2^(12/7.1) (not much different from An, is it?)
Just before the nth dose is given, the first dose has been metabolizing for 12(n-1) hrs, the second dose for 12(n-2)hrs, and the (n-1)st for 12(n - (n-1)) hrs. The total concentration Pn then is
Pn = A1 (1/2)^(12(n-1) / T) + A1 (1/2)^(12(n-2) / T) + A1 (1/2)^(12(n-3) / T) + ... + A1 (1/2)^(12(n - (n-1)) / T)
= A1 [ R^1 + R^2 + ... + R^(n-1) ] where R = (1/2)^(12/7.1)
= A1 [ 1 - R^n ] / [ 1 - R ]
D) For large n and R < 1, R^n ~ 0, so Pinfinity = A1 / (1 - R) is the minimum amount in the pt.
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