Hi, my name is Rebecca Muller. Exponential and logarithmic functions are used to model some naturally occurring phenomena. We're going to look at some of those applications during today's session. We're going to begin with exponential growth and decay, move on to the logistic growth model, look at the Richter scale, and then look at computing pH using the acidity model. So let's see how this connects to things that you've already learned about. You may recall that when we first started exponential functions, we used some interest formulas. And the one I'm interested in now is the one that had do with interest compounded continuously. So the formula looked like this. It was A equals P times e raised to the rt power. What I'm going to do is use that same formula, and just do a little change in variable to make it more generalizable. So instead of A, I'm going to call it just a y value. The P, I'm going to use y sub 0. e is a number, so it's still going to be e. Instead of r, I'm going to use k, and then I'll have t. So let's just redefine those variables in this format. The y is going to equal the amount we end up with. The y sub 0 is the initial amount, or the amount that we begin with. The value of k is still going to be our rate, but the reason I've changed it instead of an r is when we think about money, we're always thinking about our rate as being a positive value. When we're talking about exponential growth, our k value is going to be positive. But when we're talking about exponential decay, our k value is going to be a negative value. So we'll be able to have both positive and negative values for our value k. Our t is still going to stand for time, but in this context, we're not going to be kept to the idea that our time has to be given in years, as we were when we were talking about money. So let's look at our first application now. The number of bacteria in a culture is increasing according to the law of exponential growth. We begin with a culture of 50 bacteria, but after three hours, there are 200 bacteria. How many bacteria will there be after 12 hours? So what I want to note is the minute you read something and it says that it's growing using the law of exponential growth, we want to understand that that's going to be the formula we just came up with. So let's go ahead and write that down to start with. We're going to have y equals y sub 0 e to the kt power. Now, I'm going to transfer from the written words the information I need in order to solve the problem. So let's continue working down. We see, OK, it's increasing according to the law of exponential growth. That's good. We've got that. We begin with a culture of 50 bacteria. So what does that tell us? Well, that's our initial value. So we know that y sub 0 is equal to 50. After three hours, there are 200 bacteria. So that tells us that if t is equal to 3, so three hours later, then we have that y is equal to 200. That's the amount we end up with after that time frame. How many bacteria would there be after 12 hours. So what we're asked to do is to find the value of y if t is equal to 12. Now, I transferred all the information that I need from the application as it reads on paper. So we're going to now concentrate on trying to solve the equation to find what we're looking for. Notice that we're not going to be able to do this in one fell swoop. For one thing, we're given a y value equal to 200 here, and we're being asked to find y down here. So it's really a two-part problem that we're going to have to deal with. We're going to use this formula, and the first thing that we're going to do is substitute in the values that are above the line that I've drawn. Let's see what that looks like. We've got our y value, which is 200-- so I can go ahead and replace my y as 200-- equals our y sub 0 value, which is 50. e is e. And notice I do not have a value for k. So I don't know what the growth factor is going to be. But I do know that my t value is equal to 3. So I can rewrite my exponent as 3 times k. We're now able to solve this exponential equation for the value k. To do so, we'll isolate the exponential part of the equation. That is, I'm going to isolate e to the 3k. And I can do that by dividing both sides of the equation by the value 50. So 200 divided by 50 is going to be 4. 4 is going to equal e to the 3k power. Now what can we do to solve for the exponent? Well, we know that we can use logarithms in order to do that. If I change this exponential equation into its logarithmic form, we would have log base e, which is natural log, of the value 4, is going to equal the exponent, which is 3k. Dividing by 3, I can end up with the natural log of 4 divided by 3. I'm going to write it as 1/3 natural log 4. So now I have my k value. And that's going to be important because in the second part of the problem, I need to use that formula once more. But this time, I need y to be the only unknown value. So let's go back to our original format and solve for y. We're going to have y equals. We're going to have our y sub 0, which is still going to be 50. Put in the e. I'm going to put in my k value, and I'm going to go ahead and put it in exactly the format that I see here, since I'm going to use a calculator anyway. So it's going to be 1/3 times the natural log of 4. And then that's going to be multiplied times our t value. And what we're interested in is finding y when t is 12, so I can put in 12 here. And now it's just a matter of using the calculator to determine what the result is going to be. So if you use your calculator, you should come up with y equals 12,800 bacteria. And notice that's showing you the important thing about exponential growth. It just happens so quickly. So in three hours, we had 200. But in 12 hours, we've got 12,800. Here's another application. We know that radioactive materials lose particles over time. This can be modeled using the exponential formula that we saw earlier with our k value being negative. Here's the problem we want to consider. A certain radioactive isotope has a half life of 1,600 years. An organic sample is found to have only 70% of the isotope it would have originally contained. How long has it been decaying? Once more I'm going to transfer over the material that I need to work with from the given word problem. So we know that we can use the idea of y equals y sub 0 e to the kt, because this is going to be an example of exponential decay. We have a half life of 1,600 years. Now, how do we use that information? Well, the half life of a radioactive material means that it takes this long of a period of time to end up with only half as much as we started with. So what that says is when I know that the half life is 1,600 years, it means I'm going to be able to substitute in here and figure out what our value for k is going to be. However, we're then told that we get an organic sample and we find out there's only 70% of the original isotope left. So let's think about this in a second step. What we have is 70% of y sub 0 left. 70% of the original amount is going to be left. And the question is, how long has it been decaying? Which means that we want to find our value for t. So now that we transferred over all the information that we need, let's see if we can work with it. So for the first part, we're going to try to find our value of k, which is going to be our rate of decay, by using the fact that the half life is 1,600 years. So if the half life is 1,600 years, how much do we have left after 1,600 years? That y value, which is the amount that remains, is now going to be half as large as it was to begin with. So I end up with 1/2 y sub 0 equals the initial amount y sub 0 e raised to the k times my time frame, which is 1,600. What do we do now? We have two unknowns. Well, it turns out that it's pretty easy, because in this case, if I divide both sides of the equation by the value y sub 0, whatever it may be, I'm going to end up without an equation with y 0 in it. So rewriting this, we end up with 1/2 equals e raised to the 1,600 k power. Now, that's an exponential equation. We're looking for the exponent. We can change it to logarithmic form. My base is e. So this is going to be the natural log of 1/2 is going to equal the exponent, which is 1,600 k. And now to solve for k, we're going to take the natural log of 1/2 I'm just going to decide to write it as a 0.5, and divide that by 1,600. On your calculator, you would end up substituting these values in in order to get an approximate value for k. And I'm going to go ahead and write this down here. What I get on my calculator looks like this. It's negative 4.332. And then you might see this-- e and then a minus 4. What that is meaning is that this is in scientific notation, and this is k equals negative 4.332 times 10 to the negative fourth power. So any time you get this kind of result, this is what it means. If I go ahead and now just change that into a decimal format, remember that when we multiply by a power of 10 that is negative, we move the decimal place that many spaces. So I'm going to end up with a negative. And I'm going to have to have three zeros and then my 4332. So I've been able to solve for my k value. Now that I have my k value, I'm going to be able to go to the second part of the problem, where we're going to try to solve for t If I know that 70% of the initial amount is left. Let's do that next. So I'm going to end up with 70%, which I can write as 0.70, of my initial amount-- that's of the y sub 0-- is going to equal the initial amount y sub 0 e raised to the k power, which is going to be the negative 0.0004332, times our unknown t, which is what we're looking for. Once more, I notice I can divide out the y sub 0's. So this ends up being 0.70 equals e to the power that we just had written before. And how do we solve for the exponent? We use logarithms. So taking the natural log of both sides of the equation or by changing it to logarithmic form, we end up with the natural log of 0.70 equals the exponent. And to solve for the exponent, which is going to be-- I have a negative there-- I'm going to divide by t-- or to solve for t, I should say, I'm going to divide by its coefficient, which will be the natural log of 0.70 divided by negative 0.0004332. Again, using a calculator is going to give us our value. And we need an approximation. And we're just going to go to the nearest year. So we're going to have 823, which means 823 years have passed for this organic sample. It's time for a quick quiz. A culture of bacteria doubles every minute. After 30 minutes, there are 1 million bacteria in the culture. When where there only 1/2 million bacteria? A. At minute 15. B. At minute 25. C. At minute 29. Choose A, B, or C. You're correct, the answer is C, At minute 29. Sorry the answer is C, At minute 29. Consider that we're just looking at having half as many bacteria as we had at 30 minutes since it doubles every minute it's going to be half of whatever it was on minute 30 at minute 29. So if there were 1 million bacteria at minute 30 there's one half million bacteria at minute 29. So let's look at a new application. This one has to do with some populations, and sometimes populations exhibit growth in a different way than we've seen earlier. What occurs is they start off exhibiting that exponential growth pattern, but then even though it's still growing, it's doing so at a declining rate, and eventually reaching some sort of a limit. This is called the logistic curve. And the equation for that is y equals A divided by 1 plus B times e to the negative rx power. So now let's look at a word problem dealing with this logistic curve. A flu virus is spreading in a region of 10,000 people, and it can be modeled by the equation y equals 10,000 divided by 1 plus 9,999 times e to the negative 3/10 times t, where t equals the number of days after the initial infection, y equals the number of people infected after day t. How many people are infected on day 10, and then when will 50% of the population be infected? So once more, I'm going to transfer over the information that I need from the word problem. So I certainly do need the equation. That's going to give us y equals-- we have the 10,000, divided by 1 plus-- we have 9,999. And then e. And this is raised to the negative 3/10 t. What else do we know from the problem? Well, we know that t equals the number of days after the initial infection. So this is our time frame. What do we know about y? y is equal to the number of people infected. And there's one other bit of information that we want to pay attention to, and that is that our original amount is the 10,000. So we have the 10,000 people that we're dealing with. And we'll have to see how that plays out in the second part of this when we're interested in 50% of the population. There are really two questions here. One is, how many people are infected on day 10? So how can we interpret that? To find the number of people infected on day 10, we have to realize that what we're trying to find is the value of y. So we want to find the value of y. And we're told that our number of days is 10. So that means when t equals 10. So that's part one of the problem. And the second question that we're given is, when will 50% of the population be infected? So the word "when" makes you think of finding time. That's going to be the variable that we're going to deal with here. So the second part is to find the value of t. And when is that going to occur? That is going to be if we know that the number of people infected, which is going to be y, is equal to 50% of the population. The population was 10,000 people, so 50% of that will end up being 5,000. So this is a two-part problem that we're going to try to solve. Let's do the first port now. Let's find the value of y when t is equal to 10. That simply means substitute into the equation for our t. So we're going to have y equals 10,000 divided by 1 plus 9,999 times e raised the negative 0.3 multiplied times 10. And so we can simplify this. y equals 10,000 divided by 1 plus 9999e. And if I multiply it by 10, I end up with negative 3. Now we're just going to use our calculators to determine what that value is. And if you are following along with me, remember that when you're putting in the calculator, make sure you put parentheses around the denominator to make sure you're dividing by the entire value here. And you should come up with a value of y which is approximately equal to 20. And so this indicates that after day 10, we would have 20 people infected. How do we work with part 2? Find the value of t if y is equal to 5,000. Well, now I'm going to go back to the formula that we were given, and I'm going to substitute in the value of 5,000 for y. So we'll have 5,000 equals the 10,000 divided by 1 plus 9999e to the negative 0.3t. And we're going to try to solve this. So there are a couple of things that we can do. One thing you might notice is that we can multiply both sides of the equation by the denominator. So that's going to give us the 5,000 multiplied times 1 plus that denominator. An we're going to go ahead and do it that way. And that's going to equal the 10,000. And then we notice we can divide everything by 5,000. So we'll end up with the 1 plus equals 10,000 divided by 5,000, which is 2. And now what we can do is go ahead and subtract 1 from both sides of the equation. 2 minus 1 is 1. And I divide both sides of the equation by 9,999. And then let's look at the fact that this is to a negative powers. So we know that when we raise something to a negative power, it means that we can throw it into the denominator to the positive power. That means it's a reciprocal. So it's 1 over e to the 3/10t equals 1 over 9,999. Well, we can notice that therefore, the denominators must be equivalent. So just transferring it over a little bit, I can rewrite this as e to the 3/10t has to equal 9,999. Changing that into its logarithmic format, the natural log of 9,999 has to equal 3/10t. And then dividing both sides by 3/10t, we get natural log of 9,999 divided by 3/10 is equal to t. We now use a calculator once more to solve for t. And it turns out that our t value is going to be approximately 31. And so remember, t is in terms of days. And that means it takes 31 days for 50% of the population to be infected. Now let's move on a new application. The Richter scale had been used in the past in order to figure out the intensity of earthquakes. It turns out that now they use a new measure for the most part. But let's look at the Richter scale in order to look at our next application. The intensity or strength of an earthquake is a measure of the severity of the shaking. The magnitude R of an earthquake of intensity I is given by the Richter scale value of R equals log of I divided by I sub 0, where I sub 0 equals the minimum intensity used for comparison. If an earthquake has an R value of 8, how much more intense is it than an earthquake of 7 on the Richter scale? So again, we're going to transfer over the information that we need. And that is mainly going to be this formula. So we have that R equals log of I divided by I sub 0. Now, the question is to compare an earthquake of intensity 8 with an earthquake of intensity 7. So we're going to compare R equals 8 to R equals 7. Well, let's see how we can do that. We're going to start off by noting that if I look at what happens when R is equal to 7, I can replace the value of R in this formula. We're going to have 7 equals log. And what I do is go ahead and use the intensity here, and I'm going to give it a new notation. I'm just going to say it's I sub a. So that's going to be the intensity of an earthquake of a Richter scale 7. So I can now say, well, OK, I can change this into exponential format by saying that this is actually 10 to the seventh power-- because notice that's common log-- equals the I sub a divided by I sub 0. And what that gives us is that the intensity for I sub a equals 10 to the seventh times I sub 0. Well, it doesn't look like we've done much, but now I'm about to do the same thing for R equal to 8, and see how we can compare them. So back to the formula, we have 8 equals log. And I'm going to now talk about the intensity of an earthquake of Richter scale 8 as I sub B, and compare it to the magnitude I sub 0 that we had talked about or the intensity of the magnitude 0 one, which is I sub 0. Changing this into its exponential format, we're going to have 10 to the eighth power equals I sub B divided by I sub 0. Multiplying through by the denominator, I end up with 10 to the eighth times I sub 0 is equal to I sub B. Now, I'm going to do a little bit of manipulation. I'm trying to compare the intensity of the Richter scale 8 earthquake to the Richter scale 7 earthquake. One thing I note that I can do here is 10 to the eighth is the same thing as 10 times 10 to the seventh. And the reason that's important is that now I can make a substitution. I know that 10 to the seventh times I sub 0 was labeled as I sub a, which remember, was the intensity of the Richter scale 7 earthquake. So what this is saying now back into words is, the intensity of the Richter scale 8 earthquake is going to be 10 times the intensity of the Richter scale 7 earthquake. And so using the logarithms and the changing into the exponential format can give us that information. It's time for another quick quiz. How much more intense is an earthquake that measures 9.5 on the Richter scale than an earthquake that measures 6.5? A. 30 times more intense. B. 300 times more intense. or C. 1000 times more intense. Use the formula for the Richter scale in order to determine your answer and then choose from A, B, and C. You're correct, the answer is C, 1000 times more intense. Sorry the answer is C, 1000 times more intense. The formula for the Richter scale states that the value 6.5 which is our smaller earthquake would have to equal log of the quotient where the numerator is going to be the intensity of that smaller earthquake and I'm going to label it I sub s over the intensity of some bass score I sub 0. We can nail changes to exponential form so we'll have 10 to the 6.5 equals the quotient I sub s over I sub 0. Solving for I sub s I can write down the I sub s the intensity of the smaller of the earthquakes is going to equal and I'm going to multiply both sides this equation by I sub 0 to give us 10 to the 6.5 multiply times I sub 0. Now if I consider the larger earthquake which is a 9.5 on the Richter scale and I do the same process the only thing is going to change will be the exponent so therefore I can write down that the intensity of the larger earthquake is going to equal 10 to the 9.5 multiply times I sub 0. Now I really want to compare what's going on with these two intensities. What I can notice is that I that 10 to the 9.5 is the same as 10 to the 3 plus 6.5 power that's multiply times I sub 0. So now this can be re-written as 10 cubed times 10 to the 6.5 times I sub 0 using a property of exponents and then notice that the last part of this 10 to the 6.5 power times I sub 0 is equivalent to the intensity of the smaller earthquake. So therefore we're going to have the intensity of the larger earthquake is going to equal 10 cubed which is 1000 times the intensity of the smaller earthquake. Hence our answer is C. This earthquake will be 1000 times more intense if it's 3 more on the Richter scale. Our last application is going to deal with pH. Here's the problem we want to consider. The pH of a solution is a measure of the solution's acidity. A pH of 7 is considered neutral, with a pH between 1 and 7 classified as acidic, and a pH between 7 and 14 as basic. The formula for pH is pH equals negative log of H plus, where H plus is the hydrogen ion concentration per liter of a solution. The question we want to answer is, find the pH when H plus equals 8.3 times 10 to the negative 5 power. Now, we're going to transfer from here the formula that we need to work with. So we know that pH is going to equal negative log, and then we have the H plus. And the question is, find the pH if you're given H plus hydrogen ion concentration is 8.3 multiplied times 10 to the negative fifth power. So all we need to do here is do a basic substitution. Our pH is going to equal negative log. And now substituting in this value will have 8.3 multiplied times 10 to the negative 5 power. Now, one thing that we can do here is use properties of logarithms to help us simplify this. We notice we're taking the log of a product. And so we can rewrite this as the negative-- and we're going to have to use parentheses now-- the sum of the logs. This is negative of log 8.3 plus log of 10 to the negative 5 power. Let's go ahead and distribute the negative sign. pH is going to equal negative log of 8.3 minus log of 10 to the negative 5 power. Now, in this last term, notice that we're being asked to find the exponent that we would have to put on 10 in order to end up with ten to the negative 5 power. This is log base B of B to a power. And you may recall that property is going to give you the exponent as your result. So our pH is going to equal negative log of 8.3, and we're going to be subtracting the negative 5. So again, rewriting it in its simplified format, we'll have negative log 8.3 plus 5. And now we could use our calculators in order to determine this. Remember that you're only taking the log of 8.3. You're not going to be able to add this together first. So on your calculator, you would find the log of 8.3, take its negative, and then add 5 to that. And the value that you should come up with for the pH in this problem is approximately 4.1. We can interpret this to say that the solution would be acidic. Now it's time for you to try some problems on your own dealing with these applications of exponential and logarithmic functions. Good luck.