A drug with a half-life of 12 hours is taken regularly everyeight hours. Assume

Question

A drug with a half-life of 12 hours is taken regularly everyeight hours. Assume that the first dose of 50-milligrams istaken at time t = 0. The amount of drug in the blood at t = 8 hoursjust prior to the second dose is,d(t)=50e^[(-ln(2)/12)(8)]=31.5mg What is the amount of drug in the blood at t = 8(n - 1) hoursjust after to the nth dose? Sketch a graph showing theamount of drug in the blood as it varies in time. What is the amount of drug in the blood at t = 8(n - 1) hoursjust after to the nth dose? Sketch a graph showing theamount of drug in the blood as it varies in time.

Explanation / Answer

You want to sum the following:        d[t], summed overthe limits t = 0, 8, 16, ..., 8(n-1) Simplify the log term and sum the geometric progression toget:       sum =100(2-2n/3-1)/(21/3-2) To test the answer plug in n=1 and you get 50mg which iscorrect. Plug in n=2 and you get 81.5 mg which is 50 + 31.5 whichis also correct. So the formula tells us the number of mg in bloodimmediately after the nth dose. Notice that if we keep taking thedrug the amount always in the blood is found from the limit asn-> which is:    lim[100(2-2n/3-1)/(21/3-2)as n->] = 100/(2-21/3) = 135.121 mg

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