The Half Life of a drug given to an average adult is 3 days. How long will it take for 95% of the original dose to be eliminated from the body of an average adult patient, assuming exponential.
behavior for the elimination?

Answers

Answer 1

It will take approximately 37.45 days for 95% of the original dose to be eliminated from the body of an average adult patient, assuming exponential behavior for the elimination.

The Half-Life of a drug given to an average adult is 3 days. It is necessary to determine the time required for 95% of the original dosage to be removed from the body of an average adult patient by using the following information:

Half-Life = 3 days

The formula to calculate the time taken for a drug to be eliminated is:

Time = Half-Life × 2n

Where n is the number of half-lives completed by the drug.

Exponential behavior of the elimination of the drug is assumed. When 95% of the original dose has been eliminated from the body, only 5% of the original dose remains.

To find the number of half-lives, use the following formula:

Remainder = Original Amount × (1/2)²n

Where,

Remainder = 0.05

(as 95% of the original dose has been eliminated)

Original Amount = 1

(100% of the original dose)

Now substitute the values in the above formula

0.05 = 1 × (1/2)²n

Solving this equation for n:

n = 4.32 half-lives

To find out the time required for 95% of the original dose to be eliminated from the body of an average adult patient, substitute the value of n in the formula for time:

Time = Half-Life × 2n

Time = 3 days × 24.32

= 37.45 days

Hence, it will take approximately 37.45 days for 95% of the original dose to be eliminated from the body of an average adult patient, assuming exponential behavior for the elimination.

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