The halflife of a certain tranquilizer in the bloodstream is

The half-life of a certain tranquilizer in the bloodstream is 20 hours. How long will it take for the drug to decay to 91% of the original dosage? Use the exponential decay model, A = A_0 e^kt, to solve. hours (Round to one decimal place as needed.)

Solution

The model for the half life of the tranquilizer is A = A₀e^kt or, A/ A₀ = e^kt , where A₀ is the initial quantity and A is the quantity after t hours . Sincethe half life of the tranquilizer is 20 hours, we have ½ = e^20k . Now, on taking natural logarithm of both the sides, we have ln (1/2) = 20k ln (e)= 20k [ as ln (e) = 1] or, 20k = -0.69314718 so that k = -0.69314718/20 = -0.034657359. Then is A/A₀ = 91/100= 0.91, we have      e^{-0.034657359t} = 0.91. On taking natural log of both the sides, we have - 0.69314718t = ln 0.91 = -0.094310679. Therefore, t = -0.094310679 / -0.034657359 = 2.72 hours= 2.7 hours approximately. Thus, the drug will take 2.7 hours to decay to 91 % of the original dosage.