Drug elimination half lives

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If a drug has a half life of, let say 72 hours, and you take it every day, how do the levels not continuously climb?

In: Biology

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Anonymous 0 Comments

A “half life” is the time it takes for the amount to drop by half. This is an exponential decay, meaning the more there is the faster it’s eliminated. After a few half lives, the amount left is relatively small compared to the next dose.

At first, the amount in your system *does* rise pretty fast. But it will eventually reach an equilibrium where the amount eliminated in a day is the same as the daily dose.

Figuring out exactly *when* you’ll get within, say, 90% of equilibrium takes some serious calculation, but figuring out *how much* the equilibrium amount is is much easier. All you have to do is figure out what percentage is eliminated each day and set that equal to your daily dose.

For example, if a drug has a half life of one day, then 50% = 0.5 of it is eliminated each day. So set “0.5 * equilibrium = daily_dose” and solve for equilibrium to find that equilibrium = 2 * daily_dose. So if you take 125 mg per day for several days, then you’ll have 2 * 125 mg = 250 mg in your system right after taking your daily dose.

As another example, you mentioned a drug with a half life of 72 hours (3 days). First we have to figure out how much is eliminated each day. The amount left after a day is cube_root(1/2) = (1/2)^(1/3) ≈ 0.8 = 80%, so the amount eliminated each day is 100% – 80% = 20%. (Sorry, I can’t explain exponents like you’re 5.)

… Now that you know 20% is eliminated each day, you know that your daily dose will be 20% of the equilibrium amount that your system stabilizes at. Thus you can calculate the equilibrium as “0.2 * equilibrium = daily_dose”, yielding “equilibrium = daily_dose / 0.2 = daily_dose * 5”. So with the 125 mg dose example above, your body would eventually have 5*125 mg = 625 mg of the drug right after your daily dose. This is a lot more than if the drug had a shorter 1-day half life, but it still stops at a point.

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