A certain drug is eliminated from the bloodstream with a half-life of 32 hours.

Question

A certain drug is eliminated from the bloodstream with a half-life of 32 hours. Suppose that a patient receives an initial dose of 20 mg of the drug at midnight. a. How much of the drug is in the patient's blood at noon later that day? b. When will the drug concentration reach 5% of its initial level? a. The reference point is midnight. If t is measured in hours, what is the exponential decay function? y(t) = 20e^-0.019254t (Use integers or decimals for any numbers in the expression. Do not round until the final answer. Then round to six decimal places as needed.)

Explanation / Answer

half line formula = 0.693/ k

where k is y(t) = 20 e(-kt)

hance 32 = 0.693/ k , k = 0.693/20 =0.02165625

y(t) = 20e(-0.02165625t)

a) drug in the patient's blood at noon so t = 12

y(12) = 20 e(-0.02165625*12) = 15.4229595 mg

b) y(t) = (5/100)20 = 1

1 = 20 e(-0.02165625*t)

t = 138.331072 hours

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