 To episode 20, this episode is on applications of exponential and logarithmic equations. And this is, by the way, math 1050, excuse me, college algebra. I'm Dennis Allison, and I teach in the mathematics department at Utah Valley State College. Let's go to the list of objectives for this episode. First of all, we'd like to continue the discussion on logarithmic scales that we introduced at the end of episode 19. So we'd like to talk about the Richter scale and decibel level of sound. Then we'll move to exponential growth, radioactive decay, and Newton's law of cooling. Those are all equations that involve exponents and logarithms. OK, first of all, logarithmic scales. At the end of episode 19, we talked about the pH of a solution. And we were saying how numbers that are generally either very, very close together or numbers that are very, very spread out on the number line can be shown closer together if you take their logarithms. Now, let me give you an example. Take, for example, the Richter scale. Now, what would you be measuring if you measured it with a Richter scale? Earthquakes. Earthquakes, exactly. Now, earthquakes can vary in intensity, and sometimes they can be millions, even billions, of times more intense than other earthquakes, which are relatively minor tremors. So the question is, how do you describe the intensity of an earthquake when it can be billions of times more intense than some other earthquake? So here's what's generally done. If you go to the green board, suppose we say s is the minimum intensity of an earthquake that could be measured using technology. So this is the minimum intensity. And suppose there is an earthquake that occurs, and its intensity is measured to be i. So this is, let's say, the actual intensity of an earthquake that is measured on some occasion. So what we do is take the ratio i over s. Now, what we were just saying is sometimes the intensity of an earthquake can be millions or perhaps billions of times stronger than the minimum earthquake. So this ratio could be a million or even a billion. So what is generally done is you take the logarithm of that, which makes the number considerably smaller. And I'm going to call that m for the magnitude of the earthquake. And this is the Richter scale measurement that you would normally hear in the press. So for example, what if we had an earthquake that was 1 million times greater than s? So i would be 1 million times s. If I substitute into that formula, look what the magnitude turns out to be. The logarithm would be the log of the ratio of 1 million s over s. Now just make sure I've explained that clearly. s represents the minimum intensity that could be detected by the equipment. And the earthquake is actually a million times greater than that. Well, this reduces to be the log of 1 million. And so a logarithm is an exponent. And so the question is, the value of this logarithm would be the exponent that you'd put on 10, base 10, to get a million. And I think that turns out to be 6. So if the Richter scale measurement is a 6, if you have, let's say, a 6.0 earthquake, then that means the earthquake was a million times greater than the minimum intensity that could be measured by the equipment. And in fact, earthquakes can be greater than that. Typical earthquakes that I've heard of recently have been in the scale of 3.4 to maybe 4.4, something like that. So they were less than a million times greater. But that does sound very intense, doesn't it? If the intensity were 3, if the Richter scale measurement were 3, it'd be 1,000 times greater than the minimum. And if the intensity or the Richter scale number were 4, it would be 10,000 times greater. And earthquakes have actually been detected up to about 8.9 for the Richter scale measurement. So that's up close to 1 billion times greater. OK, let me ask you a question about this then. Suppose we have, let's see, I better write my formula down up here at the top. m is equal to the log of i over s. So suppose we measure an earthquake and its intensity turns out to be 855,000 times s times that minimum intensity. So what would the Richter scale number for this be? Well, if I substitute that in for i, just like I did in the previous problem, I would get the logarithm of 855,000 times s all over s. But the only difficulty this time is we're going to have to use a calculator to determine this number because I don't think we're going to end up with something quite as easily evaluated as the log of 1 million was a moment ago. So this would be the log of 855,000 because the s's have canceled out back here. And I see everybody's got the calculators out there. Stephen, what did you get for that value? 5.93. A 5.93 is the magnitude. You notice this number is almost 6. And you see if it were 6, it would have been 1 million exactly. This is a little bit under that. So we're going to get a number a little under 6. But if the magnitude had been 5, that would have been the magnitude when the intensity was 100,000 since 10 to the fifth power is 100,000. So this would be somewhere between 5 and 6, closer to 6 because this number is closer to a million. OK, so that would be the Richter scale value. So the next time you hear of an earthquake and they give you the intensity as the Richter scale magnitude, this is how you determine how it compares with the minimum intensity that you could have detected at all. Now, there's a similar situation for measuring the decibel level of sound. And the formula for decibel level, if we go to the next graphic, I think we can see it on here. This is a summary of actually the three logarithmic scales that we've talked about in the previous episode in today. In fact, let's just go up to the top. There's our pH formula for a solution. The pH is equal to the negative log of the hydrogen ion concentration. And we said that if the pH were less than 7, it would be a CDK. And if the pH were greater than 7, it said to be basic. Then there's the Richter scale magnitude, M. M is the log of i over s. We just talked about that one. And now we come to the decibel level of sound. And the decibel level, we'll call it, actually, it should be db, not just b, but the decibel level db of sound is, and that should be 10 times the log of i over i0. So there are a couple typos there, but let me just correct that formula here on the green screen. So the decibel level, db, is equal to 10 times the log of i over i0. So there was a d omitted, I think, on that graphic. And there was a 10 omitted in the formula. But if you look on the website, I think you'll see this typed in in this form. Well, once again, if you're measuring sound, i sub 0 represents the minimum intensity of sound that can be detected. In fact, that would be so soft that you probably wouldn't be able to hear it. But it could be measured by an instrument. So i sub 0 represents the minimum intensity of a sound that could be detected. Now, you know, this value was actually determined some time ago, and modern electronic equipment can detect sounds even more faint than this, but they still use this basic number here. And i represents the actual intensity of the sound that you're measuring. So for example, what if you hear a car backfire, or you hear somebody beat in a drum, or you hear somebody whisper in your ear? That would be the intensity of the sound that you hear. And that is measured numerically using electronic instruments. So then what you do is you take the ratio of i over i naught. So i would be some multiple of i naught. You cancel off the i naughts and take the logarithm. And then you multiply by 10. The reason you multiply by 10 now is because these numbers tend to be rather close together. So 10 helps to sort of magnify the numbers and stretch them out a little bit. So suppose we have a sound whose intensity turns out to be 100 times the minimum intensity that could be detected. OK, now that would still actually be a very soft sound. And so the decibel level of this sound would be 10 times the log of, let's see, i over i naught. That'll be 100 i naught over i naught, or i sub 0. And so this would be 10 times the log of 100 because the i naughts cancel off. Now what is the log of 100? It's 2, right. This is base 10. And 10 to the second power is 100. And the logarithm is an exponent. So this is actually 10 times 2. So the decibel level turns out to be 20. If I hadn't multiplied by 10, then obviously the answer would have just been 2. And if you don't multiply by 10, these numbers tend to be rather close together. So the purpose of the 10 in front is merely to kind of magnify them a bit and spread them out so they're a little bit further apart. So this turns out to be 20 decibel units for its intensity. OK, so those are the other two logarithmic scales I wanted to discuss that we had to move over from the previous episode, the Richter scale and the decibel level of sound. Let's now move to a totally different topic here and look at applications of exponential equations and logarithms to real life problems that we haven't really gotten to consider in much detail up until now. Let's go to the first graphic. This is the law of exponential growth. And it says the population n of t at time t is given by the formula n of t equals n sub 0 e to the rt power, where n sub 0 is the initial population and r is the rate of growth. Now this formula should sort of sound a bit familiar. We talked about something like this earlier when we were discussing compound interest. And who remembers what type of compounding used an exponential equation, something like this? Anybody recall? Continuous. It was continuous compounding. Very good. Yeah, so rather than having money compounded daily or money compounded in a bank account monthly or even annually, the money was being compounded continuously. That is, theoretically, every moment you wait to withdraw money from your account, you actually have a little bit more money in that account. So it's being compounded continuously. Well, populations grow in the same way. And I think we looked at an example or two of population growth as examples of continuous compounding. And what we mean by that is people are being added to the population. People are leaving the population. People die. People might move away from the city and change the population. And these sorts of things can happen during the day, at night, whatever. People are born. People die and so forth. So the population tends to be considered to change in a continuous manner. So it's not surprising that we would use an exponential function, just as we did for continuous compounding for money to represent continuous compounding for a population. Now, the difference is, if you go to the green screen, what I wrote the formula for continuously compounded money, we wrote a equals p, e to the rt power. p was the initial principal, the amount of money that you put into the account. And a was the amount in the account after t periods, or t intervals of time. And r was the rate of growth. This time, I'm writing the formula n of t equals n sub 0, e to the rt power. You notice how similar they look. p was the initial principal, n sub 0 is the initial population when you begin to take your measurements. r is the rate of growth, t is the number of years, and n of t will be the population after t years. For example, look what happens if I substitute in a 0 right here, n at 0. I would get n sub 0, e to the r times 0 power. Well, that's going to be e to the 0 power, which is 1. So this ends up being n sub 0. So the population at time 0 is n sub 0. Yeah, so that's exactly what we would have expected. Now, given more information, we can make predictions on the population at other times. Let's go to the first example. Let's go to the next graphic. In this example, it says the population of a nation is 4,200,000 and grows exponentially at an annual rate of 1.75%. What should the population be in eight years? Now, if I'm not mistaken back in an earlier episode for this section, we did a problem very much like this. So this is sort of a repeat of an earlier example. What should be the population after eight years? But then there's a second question that we've never done before, and that is, how long should it take to reach a population of 6 million people? Well, let's try solving both of those here on the green screen. See, what do we know? We know that the initial population, n sub 0, was 4.2 million. And so I think when I go to substitute this into my formula, I'll just put in 4.2 because that'll make it look a little shorter. And I'll remember that I'm computing these things in millions. Then we knew that the annual rate of growth was 1.75%, 1.75%. What would that be as a decimal? 0. what? 1.75%. 0.0175? 0.0175, thank you very much, exactly. So in other words, rather than putting in 1.75 for r, that's the percentage, we have to convert it to a decimal to substitute that in. So substituting those into my formula, we have the n of t equals 4.2 million times e to the 0.0175 times t power. Now you might say, Dennis, that's a very small exponent. Look at that small decimal you have there. This can't grow very rapidly. Well, let's see. The question in part a was to find out what's the population after eight years? So because this is the population in year t equals zero, I'll substitute in an eight, and n at eight is 4.2 e to the 0.0175 times eight power, times eight power. Now on my calculator, here's what I would do. If I were using my ti82, I would enter 4.2, and then I would press the button e to the x power, and then I would enter open parentheses, 0.0175 times eight closed parentheses, and then I would push an equal sign. By the way, when you close parentheses, what you'll see on the screen is a total of what's computed in the parentheses. So you have to hit the equal sign to go back and enter the, or to compute the entire amount. Has anybody computed that? I thought maybe some of you had computed that at your seat, so I didn't ask you to. I'll just, oh what did you get? 4.83. 4.83, okay, so Susan gets 4.83, that'll be in millions. So if in the year t equals zero, the population is 4.2 million, let's say that if the population right now is 4.2 million, then in eight years, let's say eight years from now, it would be 4.83 million. So it's increased by about 600,000, a little over 600,000. That's because the rate of increase here is relatively low. Okay, now the second question, I'm gonna leave that formula up there and go to question b, was when will the population be 6 million? Well, we've never been able to solve a problem like this before. I think part a, we had actually worked a problem like that back in episode 18 or 19, something like that. But now what do we do if we wanna find out when is the population gonna be 6 million? Well, that means I want this to be 6 million right here, the n of t. Can we go to the green screen? Okay, so the question is, when will the population be 6 million? So I'm gonna put a six for the n of t, not 6 million, cause we didn't make it 4.2 million either. We're just going with the coefficients there. So six equals 4.2 times e to the 0.0175 t power. Now, you know, in order to solve for t, I need to get the t out of the exponent. That's probably gonna mean we'll use logarithms here, but we have this coefficient of 4.2, so I think I'll divide that out first. Six divided by 4.2 equals e to the 0.0175 t power. Okay, now, to remove an exponent, I'll need to take a log on both sides. Now, I could take log base 10, but what would be a more appropriate logarithm based to use in this situation? The natural log. Natural logarithm, because we have a base e here. You know, someone may say, Dennis, should you cancel off a six, because six will divide into 4.2 and six will divide into the six, obviously. Not really necessary, you can reduce it if you want, but I'm not sure what you're gonna do if you want, but I don't think that's really gonna save us much time, so I think I'll just leave it in that form. And I'll take the natural log of six over 4.2 and that's equal to the natural log of e to the 0.0175 t power. Now, this first expression I could write as the natural log of six minus the natural log of 4.2. That's using the second property of the four basic laws of logarithms. And on the other side, if I bring the exponent out in front, I have 0.0175 t out in front of the natural log of e. What is the natural log of e, by the way? One. It's one, yeah. So this is just a one right here. So what we're down to is the difference of these logarithms is equal to this decimal times t. So t is equal to the natural log of six minus the natural log of 4.2 all over 0.0175. And that is approximately, let's see, has anybody computed that? Okay, let me just compute it right here. What I get is, I'll work it over here on the side of the screen, the natural log of six minus the natural log of 4.2. I'll enter that and then I'll divide it by 0.0175. And I get about 20.4 rounding up 20.4. Now that would be in years. And so it's gonna take the population a little over 20 years to get up to six million people. You remember just a moment ago in part a, it took eight years to go from 4.2 to 4.83. Whoops, 4.83. And it takes about 20 years to go from 4.2 up to six million people. So that sounds about right under these conditions that it would take about 20 years. Okay, that's a population problem. Let's see how much time we have. I think I'll do one more example of a population problem. This time rather than the population of a city, let's say we make this a biology problem. And what if I have a sealed culture? So let me just draw something like this on the screen. Suppose we have like a Petri dish and there's a certain amount of bacteria that's growing inside the Petri dish. And let's say we, since we can't really count the bacteria one by one, I mean, they're just too small. What we do is we look at the population in terms of area, how much area is covered by the bacterium. So let's say that initially time t equals zero. Let's say the population was estimated to be 3,000 units of bacteria. Now they may not actually be 3,000 bacteria, but maybe 3,000 square units of bacteria in a small grid. And if that's the initial population, let's say that after two days, I'll call this T sub two, what if the population has grown to 10,000? Now, I mean, I'm not a biologist, so this may be too short of a time, too much of a time for a population to go from 3,000 to 10,000. But the question might be, what would be the population of bacteria after, let's say now that was two days going from 3,000 to 10,000. So let's say what would be the population after five days? What's the population after five days? Well, let's see, we're assuming the law of natural exponential growth. So that means I'm assuming that the number of bacteria is equal to m sub zero times e to the r t power. And we know what t sub zero is, it's 3,000. So n of t is equal to 3,000 times e to the r times t power. Okay, now we know that after two days, the population becomes 10,000. So if I put in a 10,000 on the left, what will I need to substitute in on the right? If the population's 10,000 after two days, how would I change what I've written on the right hand side? I think I'll just put in a two for the t, because this is after two days. So this would be 3,000 times e to the r times two power. So, let's see, in this equation, it looks like we could solve for r. And to solve for r, I'm gonna divide by 3,000. So 10,000 over 3,000 is equal to e to the two r power. I can cancel off the thousands and just say this is 10 thirds equals e to the two r power. And if I take a natural log on both sides, that says the natural log of 10 thirds is equal to the natural log of e to the two r. Can anyone tell me right off hand what is the natural log of e to the two r power? It's two r, exactly, that would be two r. And therefore r is equal to, can I squeeze, I think I can squeeze this in on the screen, r is equal to one half of the natural log of 10 thirds. Now the reason I'm writing it that way is because I wanna go back and substitute in for r up here in my original formula. And this says n of t equals, let's say n naught was zero, was 3,000 rather. So 3,000 times e to the one half ln of 10 thirds times t. Well if I take r times, here's r times t, that's gonna be t over two ln 10 thirds. Now the reason I haven't evaluated the natural log, like I haven't evaluated the natural log down here, is because I can actually reduce this expression. Look what happens now. I'll try to write this a little larger so you can see it better. n of t is 3,000 times e to the natural log of 10 thirds and then outside I'm gonna put t over two. Is it clear what I'm doing there? I'm taking a product in the exponent and I'm putting one factor inside on the e and I'm putting one factor outside on the t over two as t over two. And so e raised to this power, natural log power, to the t over two power is e to the product. Now I can reduce that to be 3,000 times, what is e to the natural log of 10 thirds? That's actually property four from our four laws of logarithms. If you take base e and raise it to a log base e of 10 thirds, you get 10 thirds. To the t over two power. So here's my formula for the population of bacteria at time t. It's the initial population times 10 thirds. It's no longer base e but 10 thirds to the t over two power. And I was able to bring in my natural logarithm on the base e to make that base change. Now the question we asked was what was the population after five days? So I'll substitute in a five here and I get 3,000 times 10 thirds to the five halves power. And let me just compute that and see what we get here. Let's see, if I put my calculator in the middle, will we be able to zoom in on that? Let's just try zooming in while I, yeah, here we go. Okay, so I'm gonna try computing 3,000 times the quantity 10 divided by three, close parentheses, because see, I want that entire fraction to be raised to the five halves power, raised to the five halves, I'll call it 2.5 to the 2.5 power, enter. And what we get here is 60,858. So we'll say it's roughly 61,000 approximately, 61,000. The reason I'm rounding off into such a small accuracy is because we can't really count the bacteria individually. So I'm just gonna say that the population's grown from 3,000 to roughly 61,000 as a rough estimate for that. Let's see if that makes sense. You notice in the first two days, the population more than tripled. So in the next two days, you'd assume that it would more than triple again. So when you get to, this was t equals zero, that was the initial time, this was after two days. So after four days, it should more than triple, it should be well over 30,000. And then in one more day, it should be up to maybe 61,000. So that sounds about right. Okay, let's go to a different application. And this one has to do with radioactive decay. Let's go to the next graphic. Okay, this is known as the law of exponential decay. It's almost identical to the law of exponential growth. It says the mass of a radioactive substance is given by the formula m of t, m for mass, m of t equals m naught e to the negative r t power. The difference is we have a negative in the exponent. And I've changed the name of the function from n to m, so that's a small change. And the reason there's a negative in the exponent is because if we go to the green screen here, you see when you have a radioactive substance, let's say that at time zero, you have this much of a radioactive substance. But over time, the radioactivity drops off. And the radioactivity gets smaller and smaller and smaller and it approaches, but theoretically, it never reaches zero. And let's say that the amount of radioactive material here was, I'll call it m sub zero for the amount of mass at time zero. And at some point, it will decay to exactly half of that right about here. This would be one half m sub zero. This is the time axis. And at some time, the amount of radioactive material will be only half of what it was initially. So this distance right here is referred to as the half-life. Now you might say, Dennis, what happens to the radioactive material? Why does it decay? Well, you see in a radioactive substance like uranium or plutonium or many other radioactive materials, what happens is these atoms that are radioactive, when they decay, what they give off are these subatomic particles, like alpha and beta particles and so forth. I'm not really a nuclear physicist to explain this probably very well. But what is left behind is a stable element. And so the radioactive material doesn't just disappear, it just becomes stable. But there's still some other radioactive atoms that are still decaying. And so over time, there become fewer and fewer and fewer of them. And the amount of time it takes for half of the material to become stable is called the half-life. Okay, so with that idea, let's go back to the Greenstein, let me just point out one more thing. With that idea, you see what happens is this curve is an exponential function, but it's decreasing rather than increasing. So this function is expressed as m of t equals m sub zero, that's a zero, e to the negative r t, and there's a negative because the function is decreasing. For population growth, the function is increasing and going up. But we have a function that's decreasing and approaching zero. Okay, if we go back to that graphic, we have a problem on that graphic that we'd like to solve. And the problem goes like this. It says a 300 milligram sample of polonium 224 has a half-life of 140 days. So the question is, how much remains after one year? Well, let me just ask the class a question about this. We have 300 milligrams to begin with and the half-life is 140 days. So after 140 days, how much radioactive polonium 240 should there be after 140 days? 150 milligrams. 150 milligrams, right, because half of it would have decayed and half of it is still radioactive. Now in another 140 days, half of that will decay and you'll drop from 150 to 75. And in another 140 days, half of the 75 will become stable and you'll be down to half of 75, which is 37.5. So you see how this keeps reducing by half every 140 days. Well, the question is, how much remains after one year? Okay, now a student might say, well Dennis couldn't you just keep reducing it every 140 days until you get to a year? But the problem is, you may not hit exactly on a year. You may go from just before one year to just after one year. So it's difficult to say how much there will be in the middle of an interval like that. So if we go to the green screen, I think we can solve this algebraically. Our formula says M of T equals M sub zero E to the negative RT. So when we talk about the law of radioactive decay, you have to remember to put a negative up there. And if you're talking about population growth, you put a positive in the exponent. Okay, now our initial amount was 300 milligrams. So I'll put a 300 right there, E to the negative RT. And we know that the half life is 140 days. Let me just make a note of that over here. Half life, 140 days. I didn't write half life very well. I hope you can read that, the half life. Well, that's gonna give me yet another value I can plug in here. Because what that means is, when T is equal to 140 days, then the mass is gonna be 150 milligrams. This was in days right here. So if I substitute 150 for the mass, I'll have to put in 140 for T. 300 E to the negative R times 140. I'll put parentheses around the negative R there. Well, you notice there's only one variable left here, and that's R. So by knowing the half life, and by knowing the initial mass, I'm able to calculate R. So that's what I wanna do next, is to solve for R. And to solve for R, the first thing I'll do is divide by 300, and that gives me a one-half, equals E to the negative 140 R. I wanna hit and multiply it out the exponent. But when the variable's in the exponent, how do I solve for it? Well, I've gotta get it on the ground, and that means use logarithms. So I think I'll take a natural log on both sides. The natural log of a half equals the natural log of E to the negative 140 R. By the way, I can bring the negative 140 R out in front times ln of E. And as we said before, the natural log of E is one, so this is negative 140 R. So ln of one-half equals negative 140 R. Solving for R, I get negative one over 140 times ln of one-half. Okay, you might say, well, that's great Dennis. How are you gonna figure out how much of this radioactive material is left after a year? Well, what I have to do is get my formula precise. So I'm gonna take this R and substitute it right back up there. And this tells me that M of T is equal to 300 times E to the negative R power. Let's see, a negative R, R's already negative. Two negatives are gonna cancel. And that's gonna be one over 140 times the natural log of one-half times T. Now that looks really messy, doesn't it? But because I have a logarithm in the exponent, I can reduce this. I'm gonna write this as 300 times E to the ln of one-half. You notice I'm putting the log and the base E together. And what's left outside is gonna be the one over 140 and the T. So I'll just call it T over 140. What is E to the natural log of one-half power going to be? One-half. Is one-half, very good. So this tells me that M of T is equal to 300 times one-half to the T over 140. This is my formula for the mass after time T. Now you see it started off looking like this expression up here. And it ends up looking like a different exponential expression with a base one-half. You say how did base E become one-half? Well, what we did was we had a natural log in the exponent and I was able to change the base because of that, Susan changed that for us. Okay, so the question was, what is gonna be the mass that's radioactive, the radioactive material after one year? Well, let me just erase some of the things in the middle and I'll use the formula here at the bottom to calculate that. Are there any questions about this before I erase any of that? Anyone here in the classroom? Okay, if not, let me just take out this middle portion here. And I think I'll move my formula up here to the top to kind of get it out of the way. So M of T is 300 times one-half to the T over 140 power. That's the formula that we've created. And we wanna find out how much is left after one year? Can anyone tell me what I would do with this to figure that out? How much is left after one year? What would T be? One year? Well, except we were working this in days. So we can't plug in one, it'll think that means one day. 360. 365. 365. So I'm gonna calculate M at 365. Yeah, I think David had a good point. It seems like you'd plug in one for one year, but our units on time are days. So this would be 300 times one-half to the 365 over 140 power. And that is, let's see, you know, I could reduce that exponent a little bit. Well, on a calculator, I don't see that it has that much effect. But if I divide by five, it looks like we're gonna get 73 over 28, 73 over 28. I don't know that it was worth the time to take to divide that out, because on a calculator, this is gonna be computed rather quickly anyway. If you can zoom in on my calculator, I'll lay it right over the top of this. And we had 300 times the quantity one-half, so I'll say 0.5, raised to the power of, and the quantity is 73 divided by 28 closed parentheses. And I get a zero, let's see, maybe I left out a multiplication sign. Let me try that again. 300 times the quantity 0.5 closed parentheses, raised to the power of 73 divided by 20. Oh, gotta, let me go back here and enter parentheses. Open parentheses, 73 divided by 28 closed parentheses equals, there we go, 49 point, oh, let's say about 49.2. So I think the problem was I didn't put in a multiplication sign after the 300. So this will be about 49.2 milligrams. Let's just compare that with what we could calculate quickly. You know, originally, when time t equals zero, we had three, when time t equals zero, we had 300 milligrams. And then when time, when t was 140, we were down to 150 milligrams, because half of it decayed. And after another 140, that puts us up at 280, this'll be 75 milligrams, because half of the 150 decays. And after another 140, that's gonna put us more than a year, 420, it's gonna be 37.5 milligrams. So we figure the answer should be between 75 and 37 and a half, and we got 49.2. So I think that seems perfectly reasonable for this problem. Okay, let's see, let's go to the next graphic and look at Newton's law of cooling. Okay, here is the third application of exponential equations and logarithms. And Newton's law of cooling refers to Isaac Newton. And here's generally what Isaac Newton says. He says, if an object, let's see, if t sub zero is the initial temperature of an object, let's say you're taking a baked potato out of the oven. So t sub zero is the temperature of the baked potato. And t sub r is the constant room temperature. So let's say the oven is in a room that's 75 degrees. So when you take the baked potato out, the potato begins to cool, but the room stays at a constant temperature. So the room temperature is t sub r. Then the difference between the temperature, capital T at little t of the object and the room temperature, t sub r, the difference between the object temperature and the room temperature is equal to the initial difference of temperatures, which is t sub zero minus t sub r, times e to the negative rt power. Now, this is another instance very similar to radioactive decay. I have a negative in the exponent because the object is approaching room temperature. So the difference is approaching zero. Let me see if I can explain this with a graph and we'll come back to that graphing in just a moment. Okay, here's the time axis. Here's the temperature axis. And let's say here's the room temperature, t sub r. That's what I call t sub r. So I'm gonna draw a little dotted line across there. And when I take the object out of the oven, let's say its temperature is way up here because it's very hot, we'll call it t sub zero. This is the initial difference, t sub zero minus t r. So that difference is t zero minus t r. However, the object begins to cool and we're assuming the room temperature never changes. And therefore the difference begins to decay, the difference in temperatures. And so at any given time, capital T of little t is the temperature of the object and this is always up here at t r. So the difference in those temperatures begins to decay and that difference approaches zero. So the way this is expressed is to say capital T at little t minus t sub r, that would be the difference between the cooling object and the room temperature is equal to the initial difference. Okay, this was the initial difference over here times e to the negative r t power. So this time I have differences that are decaying, not the actual temperature, but the differences are decaying and this formula will tell me what is the difference in temperatures at any given time. Now if I solve for the temperature at time t, I just need to add t sub r on the other side. And so this formula looks like this, where I put t sub r added on the other side. And if we go back to the graphic, you'll see the second formula is also expressed on that graphic. Can we go back, there we are. So the first formula I wrote says that the instantaneous difference between the temperatures is equal to the initial difference times e to the negative r t power. R expressed in another form, capital T at t equals t zero minus t r e to the negative r t plus the room temperature. Now this is a bit of an idealized problem because if you take a hot object like a baked potato and set it in the room, it actually does change the room temperature ever so slightly. So the room temperature doesn't stay constant. But if we assume this is a big room, so there's a lot of air that would have to be warmed up. And let's say the air conditioning's running and so the heat is being blown away and the room temperature stays constant, then under those conditions, Newton's law of cooling applies. Okay, so we have an example of this. The example says a cup of coffee with temperature 200 degrees Fahrenheit is placed in a 70 degree room. After 10 minutes, the coffee temperature is 150 degrees. Find the temperature five minutes later. Okay, well let's take our formula. So first let me express my formula up here. The room or the object temperature is the initial difference in temperatures times e to the negative RT plus TR. That was because I moved it over from the other side. And the only reason I'm writing it this way is because it allows me to isolate the temperature of the object. Now, the initial object temperature was 200 degrees. And the room temperature, which we're assuming remains constant, is 70 degrees. And we know that the temperature of the object after 10 minutes was 150 degrees. Okay, what this is gonna allow me to do is to calculate R. And once I calculate R, and I know all the other numbers in here, then I can calculate the temperature at any given time. Okay, well let's just assemble this information into our formula. The temperature of the object at time t is the initial difference. Well, let's see, what's the initial difference in temperatures? 130. 130, so I'll put 130 there. e to the negative RT plus, then the room temperature is 70. Okay, now, that's my basic formula, but I don't know the value of R. So now I'm gonna substitute in the 10. And after 10 minutes, the temperature is 150 degrees, and I'll put in a 10 for t. So I get 130 e to the negative R times 10 plus 70. You notice this is an exponential equation, and there's only one variable, R, so I should be able to solve for it. Let's see, I'll subtract off that 70, and this gives me 80 equals 130 e to the negative 10 R power. Well, as before, I'll divide by 130, and I get eight over 13 equals e to the negative 10 R power. I'll take a natural logarithm of both sides, and the natural log of eight over 13 equals negative 10 R. I'm leaving out some steps here. You remember the natural log of e to this power is the exponent. And what this tells me is that R is equal to negative 1 tenth ln eight over 13. Hey, once I know R, I go back up here and I substitute it in. So let's write our formula over here. T of little t is 130 times e to the, let's see, now that says negative R. Well, the negative of R is a positive 1 tenth. I'll say t over 10, because there's a t out here I have to multiply by, times ln of eight over 13. Well, just as I've done in previous examples, what I need to do is get the natural logarithm with the base e. So this is 130 e to the natural log of eight over 13 raised to the power of t over 10. Let me write that a little bit better there. That's over a 13. Okay, here's the most important step. What is e to the natural log of eight, of eight thirteenths? Eight thirteenths. Eight thirteenths, okay, that allows me to make this formula look almost pretty. I know, I'm exaggerating. But it certainly looks a lot better than it did before. Okay, this is our formula for temperature. The question was, what's the temperature five minutes later? Well, five minutes after 10 minutes, that's gonna be after 15 minutes. So after 15 minutes, the temperature will be, oh, you know what, I forgot to add on my 75. There was, oh, no, 70. I forgot to add my room temperature on the end of that. I left off that guy. So after 15 minutes, we get 130 times eight over 13 to the 15 over 10 power plus 70. Well, of course, that's the three halves power. Let's go to the calculator and see how much this is. I'll put my calculator in the middle of this and we'll turn it on also. And so I need to take 130 times the quantity eight divided by 13. And I need to raise it to the 1.5 power. You know, we had 15 tenths, that's 1.5 power. And then I need to add on 70. And the temperature of the cup of coffee should be 132.7. So yeah, we'll say 132.8. So this is approximately 132.8 degrees Fahrenheit. Okay, so theoretically, that's what we would determine the temperature to be. Now you might say, well, Dennis, is that actually the temperature of the cup of coffee? Well, you know, in the real world, when you put a cup of coffee in a room, the coffee cools down, but also the room heats up ever so slightly. This doesn't take into account the heating of the room as well as the cooling of the coffee. So if the room stays at constant temperature, this should be a fairly accurate answer. But in real circumstances, this may be off by a degree or two, something like that. So, but this is all idealized. We have one more example I'd like to look at before we run out of time. This one's sort of exciting. Murder, they said. Okay, here's the situation. You've found a body. We assume the original temperature was 98.6. But when you find the body, when the police find the body, the temperature is 90 degrees Fahrenheit. And you've arrived at 9 o'clock. That's when you took this 90-degree temperature. So what you do is take another reading a little later. The temperature 15 minutes later is 88.8 degrees Fahrenheit. And the room has stayed 70 degrees Fahrenheit. When did the murder take place? Okay, so in this case, I'm going to let T sub zero not be the initial temperature of the body, but the temperature when the temperature was first taken, that was 90 degrees. So this would be at 9 o'clock. And at 9 o'clock, the temperature of the body was 90. The room temperature was 70. And I'll have to put a 70 over here. Now, we know that 15 minutes later, let's say 15 minutes later, well, we know that the temperature 15 minutes later was 88.8. So I'll put an 88.8 here. And I'll put in 15 minutes for T. That'll be 20E to the negative 15T plus 70. So 15, whoops, times R. So 15 minutes later, the temperature is 88.8. Now, what this does is allows me to solve for R. I'll subtract off a 70 and get 18.8 equals 20E to the negative 15R. And then as you've seen me do before, I'll divide by the 20. The 0.94 equals E to the negative 15R. Taking a natural logarithm on both sides, the natural log of 0.94 is negative 15T. I've left out the step where I bring the exponent out in front. It's just negative 15T. 15R. I'm sorry, there should be R's there on both of those. So solving for R, that's going to be negative 115th ln 0.94. Okay, now, with that, I go back to my original formula up here and I substitute for R. And now I have a clear definition of the temperature of the body. So this says the temperature at time T is 20 times E to the T over 15 ln 0.94 plus 70. I've done a little manipulation here. I've multiplied the T into the numerator. So this becomes the temperature at time T is 20 times E to the ln 0.94 power raised to the T over 15 plus 70. Now, what does this reduce to be here? E to the natural log of 0.94. It becomes 0.94. 0.94, exactly. So this is 20 times 0.94 to the T over 15 power plus 70. Okay, this is our definition for the temperature of the body at time T. We want to know when is the temperature 98.6. So I'll put 98.6 here and I'll solve for T. T is going to be a negative value because we're talking about times prior to time T equals 0. That was 9 o'clock. So I'm going to solve this for T. 0.94 to the T over 15 power plus 70. Well, to solve for T, I have to use logarithms again. I'll subtract off the 70 and I get 28.6 and the next step is divide by the 20. So I get 1.43 equals 0.94 to the T over 15 power. I'm going to take a logarithm on both sides. So I have ln of 1.43 equals T over 15 times ln of 0.94. I need a little room so let me move over to this other side here and we'll just chop that portion off there. So T is going to be 15 ln 1.43 divided by ln 0.94. Now how much is that approximately? Well, let's do this on the calculator. So if you can zoom in on this, we're going to get 15 times the natural log of 1.43 divided by the natural log of 0.94 and that's negative 86. We'll say negative 87. How about negative 90? That's an hour and a half, negative 90. So what we get here is approximately negative 90 minutes. So that says the murder was committed about an hour and a half before 9 o'clock so we figure the murder was at 7.30 p.m. More or less. Of course there's room for error but that gives us a rough estimate. Next time you're watching Law and Order you'll know how they figure these things out. Hey, I'll see you next time. We're going to be reviewing for exam number three. That's an episode 21. And you'll be able to use a calculator on some portions of this next exam for problems like this. We'll talk about this in the next episode and I will see you then.