 In this video, I want to offer a slightly different application of exponential growth. In fact, I won't even call it growth at all. This is an example of exponential decay. So this is a common application from chemistry where if you have some radioactive substance, so you have some big chunk of something like maybe you have your plutonium to run your DeLorean or something back to 1955 or whenever, it's a radioactive substance. So what that means is particles are being emitted out of it as as atoms fall apart, like neutrons and protons and electrons separated from each other, whatever. You basically your particles are falling apart and then these particles fly out into space, right? That's what makes radiation so dangerous for us humans that as those particles run through our bodies, they can damage our cells and things like that. We don't know. You don't have to deal with that. But as particles are lost, these little bits like these little electrons, these little protons, right? It's a teeny bit of mass, but over time, the mass of this object is going down. So you start off with some mass, but then it decays with respect to time and it actually then approaches zero, right? And so this is an example of exponential decay. Things are getting smaller over time. And so in this example, I want to show you how you can use our exponential decay model, which is really just like the growth model to model how, say, a radioactive isotope might lose mass over time. So if we take, for example, Polonium 210, so this is an isotope of our radioactive element right here, let's say that it's half life is given as 140 days. Now, some of you might realize that half life was this really old video game. First person shooter, I believe, actually, I confess I never played it. But anyways, it might, it could be an RPG for all I know, but anyways, I'm not talking about the video game. I'm talking about, in chemistry, half life of a radioactive isotope is the time required for half of the mass of this radioactive substance to decay. So this is going to be a measurement of how quickly it's decaying. And so suppose we have a sample of the substance of 300 milligrams. What I first want to do is come up with a function. Is there a formula that can model the decay of this thing? Now there's two kind of approaches that one could take to this. So the standard one is usually, they say something like the following, the amount equals the initial amount times e to the rt. And then you kind of start working through the numbers there where a zero is the initial amount. So we see that's going to be 300 milligrams. Okay. And then the rate, we don't know the rate, but we do know the half life. So the thing is, after 140 days, we see that the amount is going to turn into half of that amount, which is 150. So you get 300 is equal to e to the r times 140, the number of days that have passed, for which then you can start solving for r to figure out what is the decay rate. Divide both sides by 300. The left hand side is going to turn into one half because it's the half life. And then you get e to the 140 r. But as this is the natural exponential, you can take the logarithm of both sides, the natural log. You're going to get the natural log of one half, which by logarithmic laws, this is the same thing as negative the natural log of two. And this will equal 140 r. Then so divide by 140, divide by 140, you get that the growth or the decay rate of our isotope will equal negative the natural log of two divided by 140. And so this is a principle that you can use in general, right? If you know the half life of your radioactive isotope, then the rate of decay is always going to equal negative the natural log of two divided by your half life, right? So that's something you can use to work through these calculations right here. And so therefore, if you make that substitution into this formula right here, okay? I want you to notice what's going to happen here. A equals A naught is equal to E to the negative natural log of two divided by H. And then there's a times by T. So we're going to play the following little trick right here. So we're going to take A zero E. I'm going to bring this negative sign inside of the logarithm. So it's going to be the natural log of one half again. And then we're going to have T over H, like so. And since we have a product of exponents, I can actually factor the exponents to get A naught times E to the natural log of one half and then raise that to the T over H. The reason I want to do that is that the natural exponential, the natural log are inverses of each other. If you take, oh, what is happening to that symbol right there? That's a two, that's a parenthesis right there. If you take E and the race to a power of the natural log, they cancel out and you end up with A equals A zero and then you're going to get one half to the T over H power, where H is the half life. So a lot of math textbooks, maybe in chemistry textbooks as well, they tell you, oh, you measure it using the natural exponential E, which you perfectly can, I'm totally fine with that. But I want you to notice that when you solve for the right, you can simplify it and you actually don't have to use the natural exponential log. You can just use one half because again, you just lose a half every time a period H is established, the half life there. And so when you compare those side by side, we have the formula, is it gone now? It was up. I never actually wrote it in there. Okay. So what we're going to do is I'm actually going to approximate negative the natural log of two over 140, that'll give you negative 0.00495, which that would give you the formula A equals 300 milligrams times E to the negative 0.00495T. Now you do want your growth rate to be negative because that suggests you actually are decaying over time. Alternatively, we can use the formula that we saw, this other one we generated, A equals 300 times one half raised to the T over 140 power. Now personally, if I had to choose, I'm going to use this one right here because this rate right here is an approximation. When you work with E since it's an irrational number, it's going to be an approximation. So because of those approximation, there's going to be error inherent to that model. This one right here, at some point I'm going to have to maybe take a fraction of power and have to do a radical of some kind, but this one on the right is inherently going to have less error in it because there's less rounding going on here. So I might ask the following question, like after say 365 days, that is after one year, how much stuff do we have left? What we're trying to figure out is we're supposed to take A of 365, sorry, no leap year. This would look like 300 times E to the negative 0.0495 times that by 365, like so, for which then we just try to compute these numbers right. If you multiply together the exponent, you take the rate times 365, that will be approximately negative 1.8, 0, sorry, 0, 6, 7, 5. You want to have a lot of decimal places here to make this thing more accurate, the more accurate, the better, right? The more decimals you have. If you take E to that power, that'll be approximately 0.164, which then we times that by 300, that's going to be approximately 49.256 milligrams, which as we're measuring mass of an object, you can have a decimal part, no problem with that whatsoever. Now I want to compare this calculation we just did a moment ago with the calculation we would get if we use the green model that is using the 1.5 right there. So instead of using the natural exponential, if we use this one right here, we would want to do A of 365, great. This is going to look like 300 times 1.5 to the 365 over 140. So the first thing to do is then to take 365 and divide it by 140. And so if you wrote that approximation, maybe like 2.607 or whatever, but I'm just going to keep that in my calculator and I'm going to raise 1.5 to that power or better yet you can just stick in the fraction 365 divided by 140 when you raise that power. And then you hit equals on your calculator at that point. That's going to give you 300 times. Notice here 0.164124, if I do four decimal, I do, if I do just six decimal places, whatever, you'll notice that if you look, if you rounded it to three decimal places, these things are the same number, right? So, you know, we're going to be pretty good here. So that if you kept on going, you times that by 300, you can see that the error I was predicting seems to be quite minimal in this example. You end up with like 49.237 is what I'm getting milligrams. So it's a little bit off. Now honestly in a scientific setting like a chemistry lab, you're probably working with significant digits. And so if you had like significant figures of only three, right, then this would be basically the same thing, although I guess they do round a little bit differently. So do be cautious of these things. So we could predict how much material we have after a certain time allotment. We could also ask the following. So when, when will it be 200 milligrams, okay? How do we work through that one? Well, if we tried the first model, when will it equal 200 milligrams? So we would set 200 equal to 300 e to the negative, oh, what was my right here? Negative 0.0495 t. So then we just start solving this equation. So divide both sides by 300. We see the left-hand side turns into two thirds e to that negative power. We need to take the natural log of both sides, oops, take the natural log so you get the natural log of two thirds, which is the same thing as the natural log of two minus the natural log of three if you care. If you get negative 0.004952, we're going to divide by that coefficient right there. We have to do it to the other side as well. So we're going to divide by r and we end up that t would equal, again, when you crunch those numbers, you'll end up with 81.9, what's our unit of time, remember that's days, it's the same way we measured the half-life. And so we can probably fairly round that up to 82. And that would then give us the expectation. After approximately 82 days, we would have lost 100 milligrams of the material. And so that does give you one of the why the base e approach is a little bit healthier in this regard, because after all, the error wasn't off by much. If you did this green approach, we would take 200 equals 300 times one-half t to the 140 power right here, not a big deal. If you wanted to, you could switch this to be base two, so you get 300 times two to the negative t over 140, that works as well, 200. Divide both sides by 300, you're going to get two-thirds again, equals two to the negative t over 140. And at this moment, you're going to have to take, you're going to have to get rid of the base two, right? So we could switch things to log base two, but one thing you could do is you could just take the natural log of both sides. That's a fun little thing to do here, because you'll notice when you do that, the left hand side is the natural log of two-thirds. I swear I've seen that before. And then by exponential rules, you can bring this out, and you're going to end up with negative t times the natural log of two over 140, huh? Negative natural log of two over 140, I feel like I've seen that somewhere before. Where did we see that? Number, oh yes, it was this rate we had before, what a shocker. In which case then, coming back to the problem you have at hand, you're going to see that the answer is going to turn out to be the same, that you end up with the natural log of two-thirds over the rate, which was, remember, negative the natural log of two of 140. So really the two approaches are going to give you the same thing. So if you want to use e in the natural log, you can do that. You can also use this one-half power, that is the power of one-half, and either of these would give you basically the same answers each and every time. And so we can use our exponential models for modeling radioactive decay.