>> Let's suppose that 20 grams are released. Twenty grams of this substance is released from this nuclear reaction. How long will it take for the radioactive iodine to decay to a level of one gram. Well, we could figure it out by saying, gee, in 60 days, if we start with 20 grams, in 60 days, we'll have 10 grams. In another 60 days, we'll have 5 grams. In another 60 days, we'll have half of that and so on. So we could kind of get at it like that, but let's make a mathematical model to, to tell us how long it's going to take for one gram to remain. Now, we start with our, our basic model. Y equals AE to the BT power. Now, notice I'm using T as the variable because we're talking about a time here. We could talk about this as an X if we want to. Also I might mention to you that, that sometimes the, instead of using a A here, a C is used here for the initial quantity, and sometimes this constant that is going to tell us about how the graph behaves is a K rather than a B, but, but it doesn't really make too much difference. Alright. It's important to know that the A, this coefficient of E is an initial quantity, and the Y is the quantity remaining after a certain period of time. That's a very important idea. Now, the first thing that we generally do in problems like this where we're using kind of information or an experiment is, is to establish the value of that B or in some cases K. But, so we're using the information, and here's how we use it. We have a half life situation involving 60 days for half life of the material. Now, in this problem, we are starting with 20 grams of the stuff, and after 60 days, we have 10. So we can kind of fill in like this if we wanted to. You see, and now, dividing on both sides by 20, we get this, and then one half equals E to the 60B, and now we would go about solving for B. But I think it's important for us to realize that there's nothing magical about using 20 and 10 because after all, if, if we start with any quantity right here, we're going to have half of that amount over here. That is, we could start with 10 and have 5 over here. We could start with 4 and have 2 over here. And in all of those circumstances, once we divide by this coefficient, we're going to get one half over here. As a matter of fact, we can start with a very general value, A, you see, and then over here, we would, we would say, OK, one half of A is remaining after 60 days, you see. And then A divides out on both sides so we have the one half over here as expected. Alright. Now, let's continue from here. We would go into the exponential world by using logarithms. We would say LN, one half. Well, I'll use .5 equals LNE to the 60B. We're solving for B now. LN, .5 then is 60B times LNE, which goes out. Dividing on both sides by 60, we find B to be LN.5 over 60. Now, it's important to realize that LN.5 is a negative value. And of course, 60 is positive, but because we have a fraction that turns out to be negative, the model, the graph of the model will descend. You see, that's why this is a decaying model because we're constantly reducing the amount of material that we have over time. Alright. Now, let's use this, this information about B to solve our problem. We rewrite our, our original equation. Now 20 is the initial quantity of material, and the B value has been established, and that allows us, then, to solve any number of questions or answer any number of questions having to do with time and quantity remaining. You see, we can make this quantity remaining anything we want to and figure the time, or we could say how much quantity is remaining after a certain amount of time. We could fill in for either one of these and solve for the other. OK. In this problem, we want to know the time associated with a quantity of 1, and so we just pop in a 1 here, and then solve for T. So let's see. We would divide on both sides by 20 to clear this coefficient. So we would have this, and then we would jump into this exponential world by using logarithms, LN on both sides, and then on the right side, this exponent becomes the coefficient, and the LNE goes out, and now we'd like to isolate the T. One way to do it is, is in two steps. We could use a one-step method, but the two-step method is to clear the fraction first, multiply on both sides by 60, and the, the fraction clears here, and we have 60 as a coefficient over on the left, and then we would divide by LN.5 on both sides, and then evaluate. Using the calculator, we would find the time to be approximately 259.3 days to have one gram of that material left.