Question

Consider a drug that is taken in 40 mg doses daily. suppose that the first dose is taken when t = 0 and that t is measured in hours.

Answer 1

There are 4 questions related to this problem:
1 If the half-life of the drug is 7.3 hours, what fraction of the drug remains in the patient after 24 hours?The amount of the drug is halved every 7.3-hour period, and 24 hours equals 24/7.3 of these halving periods. 
So the portion of the drug left over after 24 hours is (1/2) ^ (24/7.3) = 0.10224 2 Write a general expression for the amount of the drug in the patient immediately after taking the nth dose of the drug
One method is to combine the residual amounts from each amount, when the nth dose arises; this will contain adding a finite geometric series
So the total amount of the drug immediately after the nth dose, in mg, is An = 40+ 40(0.10224) + 40(0.10224)^2 + ... + 40(0.10244)^(n-1) 
An = 40[1 - (0.010224)^n]/(1 - 0.10224) 
3 Write a broad expression for the quantity of the drug in the patient directly before taking the nth dose of the drug
Pn = An – 40
= 40(0.10224) + 40(0.10224)^2 + ... + 40(0.10244)^(n-1) 
= 40(0.10224) [1 - (0.10224)^(n-1)]/(1 – 0.10224) 
= 4.0895 [1 - (0.10224)^(n-1)]
4 What is the long-term minimum amount of drug in the patient?
= lim n-->infinity of Pn 
= lim n-->infinity of 4.0895[1 - (0.10224)^(n-1)] 
= 4.0895(1 - 0) 
= 4.0895 mg.