 Hello and welcome to another screencast about exponential applications. Today we're going to be talking about half-life. And this problem says to observe the functioning of organs, doctors often inject small amounts of radioactive drugs into the patient before a CT scan. Iodine-123 is commonly used because it has a half-life of 13.2 hours. If a patient is given an initial dose of 25 milliliters of Iodine-123, how much will be in their system after 48 hours? Okay, so just like we did on the last problem, I think it'll be smart to go ahead and make a table. So we'll call it T and then we can use P again for population or the amount of Iodine, whatever variable, doesn't really matter. But we'll stick with P since that's what your book always uses. So again, let's pick off some information here. We've got an initial dose of 25 milliliters. Okay, so initial means our time is zero and we've got 25. Okay, now, see the next piece of information, they say it's got a half-life of 13.2 hours. So what does that mean? Well, that means after 13.2 hours, the amount we started with will be cut in half. That's what half-life means. So if we do 13.2 hours, we're going to have half of 25, which is 12.5. Okay, again, you can keep on going. So another 13.2 hours would be, oh gosh, what would that be? Let's say 26.4. And then we could take half of the 12.5 and that would give us 6.25. Okay, so these dosimals are just going to keep getting messier and messier. And obviously, we're not going to hit the 48 hours exactly that we're looking for in the problem. So we need to do this with an equation. We're going to do just like we did in the last one. So use that same equation, p of t is p0, or p sub 0, e to the k times t. Okay, now some texts will put a negative in here since we know it's to k. I prefer just to leave it positive because then when we go to solve for k, it's going to end up being a negative value anyway. So to me, that makes a lot more sense. Okay, it's going to plug in what we know then. So p of t is an amount after a certain amount of time. So that's our 12.5 equals our p0 is our initial value. So that's 25. e to the k, we don't know. That's what we've got to figure out. And then time we know is 13.2 hours. Okay, because that's what they gave us. And again, I want to check the units in the problem. So we've got 13.2 hours. We've got 48 hours, so they match. So we don't have to worry about making any changes there. Okay, so get rid of the coefficient. Let's go ahead and divide both sides by 25. Get rid of anything we can get rid of first. And that gives us a half, big surprise, half life. Equals e to the k times 13.2. Okay, we need to undo our e function. So we need to throw in a natural log. So those will wipe out. So we'll end up with a natural log of a half, which is just some crazy decimal. Equals k times 13.2. Go ahead and divide both sides by 13.2. Cut those out. So then when I punch this in my calculator, I get a value of about negative 0.0525. Okay, so you guys might want to make sure you, when you punch that in, you get the same thing that I do. Okay? All right, now we still haven't answered the question because we want to know how much will be in their system after 48. So what that means then is, let's see, we've got our p of t is p not e to the kt. Go ahead and use that equation again. p of t, that's what we're actually trying to find this time. Okay, so we'll leave that one alone. Our p not, our initial population is still 25. e to the k is what we just found. So that 0.0525 times 48 because we wanted 48 hours. Okay, so when you crunch this all in, you end up getting about 2.01 milliliters. Okay, which if we would have continued on our table, we would have seen that that was very close to what we probably would have gotten because 48 is about a little bit more than three, but a little bit less than four half-lives. So then anyway, you end up getting something that's pretty close to this. All right, there are our credits. So thank you for watching.