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How do you calculate daily dose of a med based on half life

  1. Mar 16, 2007 #1
    Assume you have a med with a 72 hour half life, but that is dosed daily. How much elimination occurs each day of the drug, and what is the steady state dose compared to the daily dose?

    Is it just a linear decrease, so you take 50/72= 0.667, and multiply 0.667x24= 16 which means everyday 16% of the drug in your system is destroyed, and you have 100/16=6 so 6 times as much as the daily dose in your system at any given time (ie if you take 10mg a day you have a steady state level of 60mg in your body)?

    If drug half life is exponential I don't know how to solve it.
  2. jcsd
  3. Mar 16, 2007 #2
    If k is the initial amount, the amount left in function of x in hours is

    R(x) = k 2^-x/72

    This is goes along the definition of a half-life: the time it takes for k to be reduced by half. In the function above, you have R(72) = k * 2^-1 = k/2. Now you can evaluate the remaining amount at any x.
  4. Mar 16, 2007 #3
    So if you take 10mg, by day 2 you are down to 7.9 mg, then 6.2mg by day 3, and 5mg by day 4.

    So in order to find the stead state dosage do you divide the day 2 value by the initial value (say 10/7.9 = 1.26582), then use 26.582% as the % that is eliminated each 24 hours? I tried it with 10, 20 & 40mg and it all works out with 26.5% being eliminated every 24 hours. If you take the value at day 2 and multiply it by 1.26582 you end up with the day 1 dosage.
  5. Mar 16, 2007 #4
    Well, (1/2)^(1/72) = 99.04%. You could say that 0.96% is lost every hour.
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