Question

a dosage of q unites of a certain drug is administered to an individual every four hours. assume that the body eliminates 75 percent of the drug after four hours. what is the concentration of the drug after 48 hours

Answer 1

The answer is 0.000005960464775390625

Each dosage is q=100% and is eliminated 75%, so after 4 hours the concentration is 25% of the original dosage.

After 4 hours the concentration is q(1-0.75)=100(1-0.75)=25%

After 8 hours the concentration is 100(1-0.75)^2=6.25%

After 12 hours the concentration is 100(1-0.75)^3=1.5625%

After 16 hours the concentration is 100(1-0.75)^4=0.390625%

After 20 hours the concentration is 100(1-0.75)^5=0.09765625%

After 24 hours the concentration is 100(1-0.75)^6=0.0244140625%

After 28 hours the concentration is 100(1-0.75)^7=0.006103515625%

After 32 hours the concentration is 100(1-0.75)^8=0.00152587890625%

After 36 hours the concentration is 100(1-0.75)^9=0.0003814697265625%

After 40 hours the concentration is 100(1-0.75)^10=0.000095367431640625%

After 44 hours the concentration is 100(1-0.75)^11=0.00002384185791015625%

After 48 hours the concentration is 100(1-0.75)^12=0.000005960464775390625%

The concentration of the drug after 48 hours is 100(1-0.75)^12=0.000005960464775390625%

So the answer is 0.000005960464775390625