Jacob K. answered • 12/14/21
McGill Grad for Nighttime Math Tutoring and Emergency Help
When I encounter problems like these, I find it more helpful to write out fully what is going on and what I'm trying to find out, and once I have my answer in words I then try to translate to mathematics.
So, one dose is two 10mg pills, giving you a total of 20mg in your system. If the half life is 150 minutes, then in 150 minutes, you only have 10mg left in your system, 300 minutes you have 5 mg left in your system, 450 mg you have 2.5mg left in your system and the first dose is no longer active. So, somewhere between 300 and 450 minutes, the dose is no longer active. What we want to do is find a mathematical formula which displays this half life relationship, and then solve that to find out at what time your body has only 4.5 mg left in your system.
This is an exponential decay function, which we know to be in the form of y=A0e^(kt), where A0=the initial value or value at time 0, k is the rate of decay, t is time and e is euler's number. What is different from exponential growth is that k will end up being negative to represent decay instead of growth. The half life depends only on k, not on the initial amount A0. So, in order to find k, we should solve the function for when y=1/2 A0.
1/2 A0=A0e^(kt) → divide both sides by A0
1/2 = e^kt → take the ln of both sides
recall that ln(1/2) = ln (2^-1) = -ln(2), as we cannot take ln(1/2) directly
substitute in to solve -ln(2)=kt → divide by k
-ln(2)/k = t.
Now, we can solve for k, given that we replace t for the given half life, which we know to be 150 minutes, or t=150
-ln(2)/150 = k
So, finally, the formula we have to model this half life is
y=20e^((-ln(2)/150)*t).
Just to confirm, plug in t=150 to see if we get the correct half life of 10, then t=300 to see if we get 5.
Now that this formula is confirmed, we want to find out what the value of t is which gives us 4.5
We have everything we need to plug in and solve here
4.5=20e^((-ln(2)/150)*t)
Divide both sides by 20, get
(4.5/20)=e^((-ln(2)/150)*t)
Take the ln of both sides
ln(4.5/20)=((-ln(2)/150)*t
multiply both sides by 150
150ln(4.5/20)=(-ln(2))*t
divide both sides by -ln(2)
(150ln(4.5/20))/(-ln(2))= t = 322.8, which is the time in minutes from the initial dose that the least amount of active agent is in the system. After this time, you would no longer expect the agent to be active in the system. 322.8 minutes / 60 = 5.38 hours out of a single dose, so I'd expect the box to round and say that you should take one dose every six hours as needed. This is to give your body some time where the dose is no longer active to see if it needs any more of the medication in the system. Six hours is expressed as t=360. So, at the time of taking the second dose, there would be 3.79 mg left, taking the second dose would then put 23.79 mg in the body and change the new intitial amount from A0=20 to A0=23.79. After 360 minutes with this, there would be 4.5mg left in the body, which is just when the dose stops having effects on the body. Adding the third dose brings it to 24.5. After 360 minutes, 4.64mg remains in the body, which is very close to when it stops having effects. Adding the fourth dose brings it to 24.64. Letting 360 minutes pass brings it to 4.67mg left in the body. Adding the 5th dose brings it to 24.67. Letting 360 minutes pass brings it to 4.67 mg again. Adding any doses beyond the 4th dose, if taken every 6 hours, will end up always at 4.67mg left in the body after six hours.
I hope that this is able to make sense for you! I tried to be extremely clear and thorough so that every part of the process can be understood. Good luck!
