Grant's girlfriend Gloria is super proud of Grant for all of his glasses viper success. But she actually has a very successful as well. She's a wildlife conservation specialist. In fact, she just got some really exciting news. She's about to go to China for a couple of months to work at a giant panda sanctuary, a nature preserve that protects pandas to help their population rebuild. Gloria needs to learn all about how pandas became an endangered species, and how effective different conservation efforts have been, so that she can be as helpful as possible during her time in China. Now, the preserve she's going to has actually made a ton of progress in helping its panda population grow. Here's a graph showing how the number of pandas in the sanctuary has changed form year to year. The horizontal axis here shows the number of years that have passed since the opening of the sanctuary, and the vertical axis shows the number of pandas living there. How many pandas did the sanctuary start out with? In other words, how many did they have after 0 years? How many did they have after 10 years? How about have 22 years? And lastly, what about after 30 years? That's this last data point which represents now.
From the graph we can see that the sanctuary started off with 14 pandas. After ten years, they had 16 pandas, which is this point. After 22 years, they had 25 pandas. And today, they have 43 pandas. That's an impressive panda growth.
We can already tell that the function graphed here behaves in an interesting way. It's always increasing, but the curve gets steeper the further to the right we move. What kind of curve is this, though? Well, to see that, we need to zoom out a bit. Here's what the graph would look like if we got rid of the domain restrictions from the story. What kind of function do you think this is? Here are your choices. Constant, linear, quadratic, cubic, piecewise, absolute value, square root, or something else?
Something else. We've never seen this kind of one-sided growth with a tail like this on the left before.
Which statement I've written here do you think properly describes the way that this panda population grew? They are protected from habitat laws, food shortages, and predators, so they have a pretty perfect happy panda life. Here are your options. The more pandas there are, the fewer are born each year. The number of pandas born each year does not depend on population size. And the more pandas there are, the more are born each year.
As the population of pandas in the preserve grows, there'll be more and more pandas born each year, that's this third choice. This makes sense. The more pandas there are, the more there will be to have baby pandas. Then, when those babies grow up, there'll be even more pandas to have even more babies. That explains why this curve is getting steeper and steeper as time goes on.
The growth that we're seeing here has a special name, exponential growth. And it's produced by what's called an exponential function. Here's the graph of an exponential function with a slightly simpler equation. I'd like you to fill out this x, f of x chart, so we can get an idea of what's going on here. I'll get it started for you with the coordinates of one point. On the graph lies the point negative 3, 1 8th. Now, please fill in the rest of the corresponding values of f of x.
Here's our chart all filled out. The f of x values you should all have filled in from top to bottom are 1 4th4, 1 half, 1, 2, 4 and 8. These numbers definitely fall in line with the kind of increasing growth that we saw with the panda population.
What can we say about these numbers in our table? Well, we said that the x values are increasing by one, as we move down the list. But, what about the values of f of x? Please to complete this sentence, to say what's going on with those numbers on the right side of the table. As x increases by one, f of x does one of these things by some number.
When x increases by 1, f of x is multiplied by 2. We can see this in the table. One eighth times 2 is one fourth, one fourth times 2 is one half, one half times 2 is 1 and so on and so forth.
Thinking about the function you just filled out the table for, let's look at what values f of x has, as x increases by 1, over and over again. 1x is 1. f of which is 2 times 2 times 2. To get f of 4, we would take f of 3 and multiply that by 2 one more time. I see a bunch of repeated multiplication here. Can you think of another way to write this? Think about things that you learned very early on in this course. Please use that more compact notation to rewrite these three values of f. Then please tell me what f of 10 is, what f of 100 is, and what f of x is.
These first three values of f we want to rewrite can be rewritten using exponents, 2 times 2 is just 2 squared. Three 2s, multiplied together, is 2 to the 3rd, and 2 times 2 times 2 times 2 is 2 to the 4th power. We can see here that the base of each of these exponential expressions is 2, and in each one, the exponent corresponds to the value of x. The same should be true for these three values of f down here. F of 10 should be 2 to the 10. F of 100 should be realizing it, you've actually just found the equation for your first exponential function. It's right here, f of x equals 2 to the x. So that you can keep acquainting yourself with different exponential functions, I'm actually going to give you a new one to check out right now.
Here it is. H of x equals 5 to the x. But what points lie on this function? Let's figure it out. Here's a table for you to fill out, so that we can organize this information well. However, in order to do this, you're probably going to need to use a calculator. Since the numbers are going to get really big, really fast. If you don't own a calculator, it's easy to find one online by searching for scientific calculator. And if you do have your own scientific calculator. It should have a button that makes evaluating exponential functions very simple. In fact, Google makes a calculator that's very easy to find and very easy to use. Here it is. It comes up right when you search scientific calculator on Google. This is what I'll be using today to do our calculations. As you can see, this calculator has a button that says x to the y on it. Now if I want to calculate six to the third power, for example, I would press 6. Then I would press x to the y. And then I would press three. You can see that the calculator is showing us 6 to the 3rd. When I press equals, that gives me the number that's equal to, which is 216. Let's try another one. Maybe I want to find four to the negative fifth. Well, I press four, then our lovely x to the y button, and then I type negative 5. Great, I see 4 to the negative 5th, which is what we're looking for. Pressing Enter tells me that it's this long exponent. Let's practice using that button on our calculator to find different points on this function. For each value of x, I've written here, please fill in the corresponding value of h of x. In other words find 5 to the x for each of these x values.
Here is our filled out table. Now, of course we don't want to be dependent on calculators for everything, but they certainly do come in handy when we have calculations that end up giving us huge numbers, like 3,125. You can see this only at a value of x equals 5. Imagine if I had given you even higher x values. Our numbers for h of x would have been huge. Some of these numbers you definitely could have figured out by hand or in your head. In fact, you probably did. Though there might be some on this list you could have done that way, that you didn't realize. For example, when x equals negative 3, h of x is over here. This answer is perfectly acceptable, as well. Although, it's probably not what a calculator will give you, depending on what sort of device you're using. We can see something similar happening for 5 to the negative 2. That's just 1 over 5 squared, or 1 over 25.
Now that you've seen a couple of exponential functions, let's talk a bit more in generally about them. We use exponential functions to model relationships in which a constant change in x creates a proportional change, a percent increase or a decrease in f of x. In other words, as we saw in our tables that we worked with, whenever we add a certain amount to x, we're looking at adding one every time in this case. f of x is going to be multiplied by a certain amount. In the case of this function, that amount was two. Now the general form of an exponential function looks like this. f of x equals a to the x power. Here a is a positive constant other than 1. Before we move on, lets just look at why a can't be 0 and can't be 1. Well, what is f of x if a equals 0? What if a equals
When a is 0, all f of x ends up being is 0 to the x. And that actually just equals 0 for every value of x except for 0, since 0 to the 0 power is not defined. When a equals 1, f of x just equals 1 to the x. And that simplifies to That's not very interesting. Horizontal lines are definitely not exponential functions. That's why a can't equal 0 or 1 for this general form of our exponential function.
We know that when we add to x, we're multiplying the value of f of x, but by how much? Well, let's consider the functions we've already looked at. In the case of f of x equals 2 to the x, f of x was multiplied by 2 whenever x increased by 1. Here, 2 is called the growth rate of the function. So what can we say about the growth rate of exponential functions in general? Considering that general function, f of x, equals a to the x, when we increase x by 1, what's going to happen to f of x?
For f of x equals a to the x. When x increases by 1, f of x is multiplied by a. A here is the growth rate. And that means we can tell what the growth rate of an exponential function will be by looking at its base. Which we can see, in this equation, is a.
There was, however, a reason that the panda population ended up as an endangered species. And that was because their population declined to a dangerously low number. More pandas were dying each year than being born since they had lost huge portions of their habitat, making food and mates difficult to find. Now, even though Gloria's Giant Panda Preserve is doing very well rehabilitating its population, pandas are actually still endangered worldwide. Preserves like hers are one of the ways that hopefully pandas will no longer be endangered anymore someday. Their overall population will one day get to a much healthier size. Which of these statements I've just written do you think makes the most sense in terms of the decline of the panda population? A constant number of pandas died each year. The larger the population, the more pandas died each year. And the smaller the population the fewer died each year. Or finally, the larger the population the fewer pandas died each year, and the smaller the population, the more died each year.
This second choice is correct. If pandas all have about the same lifetime, then they'll die off at a constant rate, not at a constant number per year. That means that the higher the population. The more deaths there will be, because a certain percent approximately, will die each year. As the population shrinks, there will be fewer pandas to die.
Here are three graphs, which I've labeled a, b, and c. And I'd like you to tell me which one fits the statement we chose in the last question. The larger the panda population, the more pandas die each year. And the smaller the population, the fewer die.
Graph C shows this pattern. We can see that as the total population of pandas, shown on the vertical axis, gets lower, the graph gets less steep, meaning that the rate of death is decreasing.
The graph that we selected in the last video is unlike any other function we've seen before. And in fact, it's a different kind of exponential function. This graph shows another function that follows the same sort of pattern. Now, since I've told you that the graph is exponential, we have an idea of what kind of behavior it exhibits. Every time we add some constant amount to x, f is going to increase by a certain percent, or is going to be multiplied by a certain number. So, for example, if we add 1 to x here, what is f of x multiplied by? I'll label a few points on this graph to make things a bit simpler. Now think carefully about this. Make sure that you fill in the number f of x is multiplied by.
Every time x increases by 1, f of x is multiplied by 1 3rd. We can see this in the points that I labeled here for you. As x steps up by ones, f of x is going from 9 to 3 to 1 to 1 3rd and so on. This is definitely different. Pretty cool.
This graph that we just came up with, is different from other exponential functions we've seen. Because it isn't growing. It's decreasing everywhere instead of increasing. We call this exponential decay instead of exponential growth. We can see that the graph of an exponential decay function looks like the opposite of an exponential growth function. It's going down instead of going up, as you move from left to right across the graph. Now decay might sound like a funny word for this sort of behavior, but it's called this because of some of the context in which functions like this pop up in nature. We would use exponential decay equations to describe a panda population going extinct, a giant group of bacteria dying off. Or the amount of medicine in the bloodstream decreasing. Instead of having a growth rate like we have in functions like this one. We have a decay rate. Thinking back to our original form of an expotential function f of x equals a to the x. What do you think that a needs to equal in order for a graph to show decay instead of growth? What interval should a lie in?
In order for f of x equals a to the x, to exhibit exponential decay. A must lie between 0 and 1, not including either of these numbers. When a is in this interval, the value of the function is multiplied by a fraction each time x increases. Making f of x get smaller and smaller. Let's experiment with this for a second. Here we are, back in friendly old Desmos, and I've typed in the general function that we've been talking about, f of x equals a to the x. I've made a into a number that I can adjust by sliding this right here. Right now, a is set to 1 and, as we calculated before, this is just a horizontal line, f of x equals 1. Now, let's see what happens as I decrease a. You can see right away, just when I let it go below 1, we already have a function showing exponential decay. As I continue to make it smaller, the decay gets more and more steep. It's happening more quickly. That's going to continue to happen until a gets too small. When a reaches 0, we have that constant function again. Except, this time, it's at f of x equals 0. We can, of course, also slide over into the exponential growth area as well. That would just mean we need to increase a past 1. Let's try that. Here were are back in our exponential decay. And there we go, we're growing. Just above 1 we already see that new pattern forming. And the growth gets steeper and steeper as a increases. Since we're going to be multiplying by more and more each time x increases. If I let a get really big, our graph's going to increase at incredibly quick rate. I would encourage you to visit the Desmos website and play around with this on your own. It's really fun to see how many different things you can do with exponential functions.
What have we learned about the behavior of exponential functions so far? Well, once again, working with our general function, f of x equals a to the x. If a is greater than 1, then f of x is increasing everywhere in this domain, showing exponential growth. If 0 is less than a is less than 1, then the function looks like this one. It's decreasing everywhere in its domain, and we say that it shows exponential decay. There is, however, another way that functions of this type are sometimes written out. I've graphed four different functions over here, and I'd like you to tell me which one is equivalent to the function we were talking about before? f of x equals 1 3rd to the x. I'd like you to think about this, using both equations I've written here and the graphs that are drawn.
The answer is this one, y equals 3 to negative x. First of all, we can to tell that the graph, this one right here, looks just like the one for 1 3rd to the x. It's the only one on this coordinate plane that's showing exponential decay. We can also see why this equation is equivalent to the one for f of x from algebra. 3 to the negative x is equal to 3 to the negative 1 to the x. 3 to the negative 1 is just equal to 1 3rd. So the equations and the graphs for these two functions are one and the same. They're the same function.
Let's talk a bit more about the behavior of these exponential decay and exponential growth functions. In particular, what are their domains and ranges? The functions that are graphed here are y equals one half to the x. Or now you know, this is the same as y equals 2 to the negative x. And over here, we have y equals 2 to the x. For each function, please use interval notation to write in the domain and range. Although using the graphs can be helpful to answer this question, the equations are also really useful. For each domain, try to see if there are any values of x that produce invalid answers when plugged into the function. And for the range, think about what values the function can and can't take on.
Both of the functions we have here output real numbers for y, for any number of x that we could plug in. So the domain of each is all real numbers. In other words, negative infinity to infinity. The ranges of both of the functions are the same as one another, as well. They are both 0 to infinity, not including either of those endpoints. Now we can see this happening on either graph, but why is it happening? Here is an instance where our equations can lend a hand. No matter how small a number you plug in for x in our exponential growth function, 2 to the x will never reach 0. The numbers will get closer and closer to 0, but they'll never actually equal it. For the exponential decay equation, if we look way out to the right side of the graph as x approaches infinity, y will approach but never touch 0, since the same thing will be happening with this equation.
Thinking about what we found in the last question, which of the following features do these exponential functions have? Please check all that are true. Y intercepts, horizontal asymptotes, vertical asymptotes and slant asymptotes.
Let's go through these possible answers one by one. To start off, both of these functions do have y intercepts. In fact, every single exponential function has a y intercept. We'll talk more about that later. Since the domain of each of these is all the real numbers, we must be able to plug in 0 for x and get a valid number for y. Which tells us where the y intercept is. Now, let's move on to asymptotes. Both graphs have horizontal asymptotes at y equals 0. Neither of them however, has a vertical asymptote or a slant asymptote. Now it's easy to mistake these steep parts of the curves for vertical asymptotes. But, that just goes to show rapidly they grow or decay. These steep sections do not represent discontinuities, or breaks in the graph, like vertical asymptotes would. In fact, since the domain is all the real numbers, these graphs are continuous.
How are the graphs of this exponential growth function and this exponential decay function related to one another? I bet you can figure it out. Once again, here we have y equals two the x, and y equals one half to the x. How do we transform one of these into the other? Do we need a horizontal shift, a vertical shift? A reflection in the x-axis, a reflection in the y-axis, or a reflection in the line y equals x?
These are reflections in the y-axis of one another. We can see that if we flip this one across this line right here, we'd end up with this exact graph. Now we can why this happens from our equations if we rewrite this one using the other form we learned. Remember, we can also write this as 2 to the negative x. A few lessons ago, we saw that if you replace x with negative x in a function. You end up with the reflection of the original function across the y-axis. That's exactly what's happening here. Turns out this other form could be pretty useful in certain situations.
Functions for exponential growth and exponential decay have some major differences. But they're also very closely connected as we've just seen. There's one small similarity that we haven't discussed yet, though. All functions of the form f of x equals a to the x, where of course, a is greater than 0 and not equal to 1. Have the same y intercept. What is it? Well, you can plug in zero for X to find out. Please put its coordinates here.
Well, to start off, 0 is the coordinate of any x-intercept. So if we plug in 0 for x in our equation, we get f of 0 equals a to the 0. Now, if I take any number that a could equal to the 0 power, then we get 1 for f. That means that the y-intercept of any exponential function of this form, whether it's growing or decaying is 0, 1. That's really cool. Bringing it back to decimals again, I've started us off with the function f of x equals 2 to the x. And we can see that, yes, in fact, our y-intercept is at 0, 1. As we slide a, either making it bigger or making it smaller, the one point that never changes is the 0.01. You can watch the rest of the curve flip right around that point. It's almost like a pivot point.
We know from the last question that the y-intercept of the function y equals 1 half to the x is 0,1 as we can see right here. What if I wanted a function identical to this one, but with a y-intercept of 0,4 instead? Which of these function do you think that is? Here's a little hint. Think about what transformation you'd have to do to move the point 0,1 to the point 0,4 instead. Then, think about what substitution you need to do in this equation to make that transformation happen.
That would be the function y equals 1 half to the x plus 3. We've seen before that in order to shift a function up by three units, we need to replace y with y minus 3. Doing that and then adding 3 to both sides, gives us this answer.
What if instead, we wanted a function identical to y equals 1 half to the x, but we wanted it to contain the point 3, 1? Remember that this function contains the point 0, 1. Which function do you think fits this description? A hint again. Think about what transformation you would need to do to move the point 0, 1 to the point 3, 1. What substitution would you need to make in this equation to perform that transformation?
This function, y equals one half to the x minus 3 power is identical to this one, but it shifted 3 units to the right. Now, I knew that shift needed to happen, because here we have the point 0,1 and to transform 0,1 to 3,1, we just need to move 3 to the right. A shift of 3 units to the right means replacing x with x minus 3, which is exactly what's happening here. Awesome. It looks like you can transform exponential functions just like you can any other kind of function. The strategies we learned, many lessons ago, are going to be just as applicable here, as they were for every other kind of function we've worked with. That's really convenient.
At the same time as Gloria has been learning about exponential growth to help her pandas out, Grant has also been trying to figure out exponential functions to help deal with his company's investments in different bank accounts. Suppose that Grant wants to invest $10,000 and the bank say, you can get 3% interest per year compounded yearly. Grant feels confused. The bank explains that compounded annually means that once a year the total amount of money in the account is calculated. And then 3% of that amount is added on top of the total in-, interest. Let's start out by just calculating this for the first three years that Grant has the account. I've already written in all the information that we have and really it's all that we need to figure out the rest of this. You know that he starts out with $10,000 at the beginning of the first year that he invests. Our interest rate is 3%. So that means that 0.03 times this amount will be added to the account at the end of the year. 0.03 times $10,000 is 300. 10,000 plus 300 makes 10,300. Since this is the amount that's present at the end of year 1, it's the amount that's present at the beginning of year fill in the missing information in each of these cells. How much interest will he earn in the second year? And how much will that give him at the end of year interest and then the new total.
Since we know how much money we're starting with the beginning of year 2. We need to figure out the interest that amount is going to earn. Since your interest rate is 3%, we need to multiply 10,300 by 0.03. Since this is how we convert three percent into a decimal. 10,300 times 0.03 is just 309. And this added to our original balance gives grant 10,609 at the end of year 2. We know that this is the amount that he starts with at the beginning of year 3. So we can write that in here as well. Now, 10,609 times 0.03 gives us $318.27. Adding these together, Grant has a total of $10,927.27, by the end of the third year.
These numbers are great but I think it would help if we looked a little bit more closely at how we just calculated each of these answers. Let's just focus in on this first here for a second. For the first year, what we found was $10,000, the amount we started out with, plus 0.03, the interest rate times that original amount of money, $10,000. Now, to help us understand in a different way what's going on here, I'd like you to factor out the 10,000 from this expression and then write something in these parentheses right here. Now, you don't need to sum anything inside the parentheses. You can just leave 2 terms in there.
If we factor out a 10,000 from each term in the original expression, then for the first term, we find 10,000 divided by 10,000 or just 1. For the second term, which we add to the first one, we get 0.03. That's our answer, 1 plus
I'd like to introduce some terminologies that we can talk more easily about the different quantities we're dealing with. We'll call the amount of money that Grant initially invested in the account, the principal, which in Grant's case is $10,000. We also already mentioned the interest rate before. And in Grant's case, this is 3%. Now, let's think back to the math that we were doing. In the expression, 10,000 times 1 plus 0.03. Where do the numbers 10,000 and 0.03 come from? For each one, please pick one of these choices. The number of years the money is invested, the number of times interest is compounded per year, the interest rate as a percentage per year, the interest rate as a decimal per year, the principal or the number of months in a year.
$10,000 was the principal for Grant. It was the amount of money that he invested initially. 0.03 represents the interest rate as a decimal per year. It looks like these two numbers are pretty important quantities.
Earlier, we rewrote Grant's balance after one year as 10,000 times 1 plus 0.03. Here, we can clearly see those two quantities we were just talking about. The principle 10,000 right here, and the interest rate, 0.03 right here. So, what would happen if we kept our expressions in this form, but found the balance after year two? First things first, though, let's remember how we found this quantity here. We took the principle, and then, since this is the amount of money in the account at the start of this year, we multiplied it by the interest rate. That's how we got this quantity. Now, we know that to find this balance after two years, we're going to need to start with the amount of money that's in the account at the beginning of year 2, which is this quantity right here. So, this amount is going to sneak around and end up as our starting point. Now, I'm about to give you a bunch of boxes to fill in, but please don't be overwhelmed. You're very well-prepared to do what I'm about ask you to do. It's not a big deal. What I would like you to do is to substitute into these green boxes a certain quantity. And I'll give you a hint, the same quantity belongs in each of these boxes. Think using this equation right here what quantity that is. Then, based on what you end up with right here, I'd like you to write this in a more compact form right here. However, keep 1 plus 0.03 written as is. Don't change it to 1.03.
To find the balance after two years, we need to first start out with the amount of money that's in the account at the beginning of the second year. We know that's the amount that the account has at the end of the first year. 10,000 times 1 plus 0.03. Now, in fact, that's the same value that belongs in these other two green boxes. The reason it goes here is because this is the quantity that we're going to use to calculate the new amount of interest that needs to be added. Now if we look at this entire expression, we can see that what can be factored out, is what's already in the green boxes, 10,000 times 1 plus 0.03. If we factor that out of each term, then what we would end up with for the first term here would just be 1, and for the second term it would be 0.03, as we see in the parentheses here. Now looking right here, what we see is repeated multiplication of 1 plus 0.03. Using exponent notation, that gives us 10,000 times the quantity 1 plus point oh 3 squared.
So now we've found the balance after one year for Grant's account and the balance after two years. What we're going to do now is something that we haven't really done in this course yet. I'm going to ask you to make a hypothesis. This just means that you're going to come up with an idea and we'll see if it's right pretty soon. This is something that mathematicians do all the time. They look at examples, look for a pattern, guess what the pattern might mean, and then test their guess to see if it's right. If it's wrong, it's not a big deal. They try again with a new guess. So the question I have for you is, what does the two here mean? Where does it come from, and what does it stand for? Think about what might be going on, in general, with these equations. And see if you can predict the pattern without working anything out. Write your guess in this box. Just so you know, this is not going to be graded.
Now that we found the balance after the first 2 years, it would be great if we could find it even further along in time. Let's work right now to find the balance after 3 years. Actually, you already did this earlier, but I'd like us to write it in this kind of notation. We start with this amount, since that's how much money there is at the start of your three. And then, the amount that we need to add to this in interest, is just this quantity times 0.03. If we factor out this quantity from both terms, then we end up, once again, with 1 plus 0.03, inside this second set of parentheses. However, since we have 1 plus this look more compact, by changing our exponent up here. Instead of squared, this is cubed this time. Now that we have another quantity continuing our pattern here, what do you think about the guess that you made just before this? Do you want to keep it as is, or do you want to change it? Think about what pattern you see. Think about what you think the exponent in each of these expressions means, and then use that information, that pattern to write down the balance in this form after 4 years, after 10 years, and after 100 years. I'll also give you another chance to write down your hypothesis. You can write in exactly what you wrote before or you can change it this time, taking into account the new information we have now. Once again, this part won't be graded. We are, however, going to see how your numbers look.
Remembering that there's an invisible 1 here, as the exponent in this expression, all of these exponents stand for the number of years that have passed. Or another way to look at this is that it's the number of times that interest has been added. After four years, then, we'll have 10,000 times 1.03 to the 4th power. After 10 years, we'll have 10,000 times 1 plus 0.03 to the hundredth power. Well, if you got all these number right, congratulations. You obviously spotted the pattern and then you extended it. That is some great detective work using math.
For this account, the pattern that emerged for all the expressions for the balances after a certain number of years, was like this. We take the principle, the amount initially invested, multiply by 1 plus the interest rate written as a decimal. And then, take that quantity, that I've written in parentheses to a power that's equal to the number of times interest has been added to the account. In this case, since interest is compounded yearly, that would just equal to the number of years that had passed. Let's try doing something similar, but with a different situation. Remember Athena? Well, she has $300 to invest. The interest rate on the account she's going to put it in is 5% per year, compounded annually. How much money will she have in the account after four years have passed? Remember that pattern that we were using before.
Athena's $300 is going to be multiplied by 1 plus the interest rate written as a decimal, so 0.05. And that multiplying is going to happen four times, since interest will be compounded four times in four years. We can, of course, use our wonderful calculator to multiply this out. 1 plus 0.05 is just 1.05. And this raised to the 4th power is this quantity. We want to multiply that by 300. That gives us 364.651875. Rounding to cents, that gives us $364.65.
The strategy we've been using to solve these problems is definitely useful, but there are different interest options for Grant to explore. What if instead of only compounding the interest once a year, his account compounded it twice a year? Now, interest rate is still quoted as a yearly rate. So let's say he keeps it at 3%. However, since his interest is going to be compounded twice a year or semi-annually, then each time the compounding happens, the rate that's used is only going to be half of this annual rate. That means that each time we do a calculation, we're not going to use the number 0.03 anymore, instead, we're going to use half of that. I'd like you to help me describe the situation by filling in some numbers for me in a few sentences. Here's what I'd like you to tell me. If Grant invests $10,000 at a 3% annual rate compounded twice a year, then in six months, he will receive 1.5% on his $10,000 investment. Which means that some amount of money, which you'll write in here, will be added to his account in interest. The total balance of the account will then be whatever you think. Then after six more months, Grant will again receive 1.5% interest, but this time, on the new amount. The interest added this time is going to be something, which will make the total amount of money in the account after one year equal to what?
If Grant invests $10,000 at a 3% annual rate compounded twice a year, then in 6 months, he'll receive 1.5% on this $10,000 investment. This means that $150.00 will be added to his account, in interest. The total balance will then be receive 1.5% interest, but now on this new amount. The interest added this time will be 0.015 times 10,150, or $152.25. This will make the total amount of money in the account after one year equal to $10,302.25. Note that in the second six months, more interest is added, since Grant is getting interest on the interest from the previous six months, as well as on his original investment. Seems like a pretty good deal to me.
As we just saw, if Grant invests $10,000 at a 3% annual rate, compounded twice a year. Then in one year, there will be $10,302.25 in his account. Now, is this more or less than Grant would have if interest was compounded only once a year?
The answer is more. As we calculated earlier, Grant would only have $10,300 if interest is compounded just one time a year. Compounding interest in this way actually gives Grant more money after a year than annual compounding would. I wonder how we could generalize about this.
We just saw that after one year, if Grant's account compounds interest twice a year, he'll end up with 10,000 times 1 plus 0.015 squared dollars. Now I want you to think back to that hypothesis you made earlier and how the situation is different this time. What does this exponent here mean? Well, remember, we're going to multiply by this number every time interest is compounded. So, what exponents belong here and here? To help us figure out how much money is in the account after 2 years and after 3 years.
We need an exponent of 4 here and an exponent of 6 here. Remember that the interest is being compounded twice a year. So to figure out how many times it has been compounded total, we multiply the number of years that have passed times the number of times compounded in a year.
Let's pretend again, for a minute, that's Grant's interest is compounded just one time a year. If you invest that same $10,000 at that same 3% annual rate, then after four years, he'll have this amount of money, as we saw before. Let's talk about what this means, this expression here. $10,000 represents the principle. This initial amount of money he invested and it goes right here. Point o three here is the interest rate written as a decimal. Three percent becomes point o three. Then last but not least we have our exponet four. This represents the number of years that the money has been invested in the account. Now this worked fine for compounding just one time a year, but it's not going to work as well for compounding twice a year. If we switch the compounding to twice a year instead of once a year, then this is the expression we came up with. So what can we say about comparing these two expressions we've just seen? I'd like you to tell me that by filling in this sentence. When we switch from compounding interest once a year to twice a year, what happens to the interest rate and what happens to the exponent here?
When we switch from compounding interest once a year to compounding it twice a year, the interest rate is divided by 2 and the exponent is multiplied by 2.
So here's what we have for compounding twice a year with Grant's situation. This of course is just an example and these numbers could be different depending on how his account was set up. However, numbers have played their roles as I've written in here. This number is his principle, the amount he first invested. Here goes the interest rate divided by 2, And the exponent is the number of years that have passed multiplied by 2. Now we always like generalizations so let's see if we can apply one here. I think we're going to need some variable names however. Let's call our principal P. r will be our annual interest rate written as a decimal. T will be the number of years that have passed since the account opened. N will be the number of times interest is compounded per year. This was two in our example up here. And A will be the total amount of money in the account. I'd like you to use these quantities we just came up with to create a general equation for A. The total amount of money in the account. Just fill in the proper letters in these different slots. Remember you can use this formula as a guideline for figuring out what goes where here.
Comparing this to what we have up here, P the principal belongs right here. Now, what goes here is the interest rate that we use every time interest is compounded. This is equal to the annual interest rate, r, divided by n, the number of times interest is compounded per year. We can see that that's what we have up here. We have a 3% annual rate. 0.03 divided by 2, which gives us this number. Great, we're really close. Now what belongs in the exponent? We know that what needs to go here is the total number of times that interest has been compounded. It's compounded n times for every t years. So, that means that the total number of compoundings is just nt. There we go, we have a formula that should be pretty easy to use. And I hope, more than anything else, that this makes sense.
We've worked really hard to come up with this brand new equation, so let's use it. Grant has actually made more than $10,000, since cleaning glasses has been so profitable. He wants to invest more money, $20,000 in another account at a quarterly which is four times a year or monthly? Well, let's help him figure that out. How much money will he end up with after 5 years in either case? This is the perfect chance to use our new equation. Please round to two decimal places.
All we need to do to solve these two problems is plug the right numbers into our equations up here. Let's start out with the quarterly compounding option. What do we know? Well, P equals 20,000, r equals 0.04, n equals 4 and t equals what we're looking for. A will equal 20,000 times 1 plus 0.04 over 4, all of that raised to the 4 times 5 power. This simplifies to 20,000 times 1.01 to the Now, on to monthly compounding. We do almost exactly the same thing. In fact, all of our information is the same, except that n is not 4. This time n is 12, since there are 12 months in 1 year. For A then, we have 20,000 times 1 plus calculator gives us $24,419.93. Although, these two total balances aren't too far off from one another, Grant will end up with slightly more money if he goes with the monthly compounding interest choice. And look at that, after 5 years in either case, he'll earn over $4,400 just by keeping his money in his account. That's a pretty good deal.
Referring back to our formula, A equals P times 1 plus r over n to the nt, we've seen n, the number of times the interest is compounded in a year, equal to several different values in our different situations. We've had n equals 1, frequently than this. As you've experienced, Grant likes to push limits in order the maximize the amount of money he makes, and now he wants to know what would happen if his interest is compounded so frequently That is actually compounded over and over again all the time. Lucky for him, this sort of interest setup actually already exists. It's called continuously compounded interest. If we want to use our good old formula here and keep increasing and increasing n Do I reach a particular function? Let's see what happens if we substitute in the values p equals 1, r equals 1 and t equals 1 and then increase the value of n. Well they just said that p equal to 1, r equal to 1 and t equal to 1. Here's what we have. A equals 1 plus 1 over n all to the nth power. Now I'd like us to see what happens if we let n increase. Let's plug in some values to find out. So here's a little table for you to help me fill out. For each value of n here, please plug it into this equation to find the value of a that goes with it. Now, you're probably going to get some pretty long decimals, so I'd like you to type in as many numbers As your calculator will give you, maybe around ten or so. Remeber you can always use the google calculator online if you need a different device.
So here are the numbers that I get for A, when I plug in n. You can see we have some pretty long decimals here, except for our first answer, which is very simple. Since we just get 1 plus 1 to the 1st power. However, you might notice, that the difference between adjacent values, here, gets smaller and smaller as n increases. The difference between these last two, doesn't show up. Until the fifth decimal place. Something interesting is definitely happening here.
As we increase n toward infinity, the number we have for A over here gets closer and closer to a certain very special number. That number is e. But what on earth is e? We haven't been using that letter to stand for anything. Well, this is a very important point. e is actually a number, an irrational number to be precise. And we can approximate it as 2.718281828459, and of course, this decimal continues forever, since this number is irrational. It's a number that has special significance for higher level mathematics, biology, chemistry and many other fields. So it's really exciting that we're going to get to work with it. Conveniently, scientific calculators also generally have buttons for e on them. On my calculator, it's right here, and when I press it, well, it just types the letter e. But if we press equals, then you see a rounded version of this number. Now, remember, this is not the exact value of e, this is an approximation.
Now that we know our value of e, or at least many digits of it. Though, not nearly an infinite number of them. We can see how those numbers we had on our table, introducing us to the idea of continually compounding interest. Definitly do approach this value. There we're getting closer and closer to it as we moved on the list as n got bigger. And in fact, the formula for calculating continuously compounded interest, does involve E. Here it is. So, for continuously compounded interest, we have this formula. A equals p times e to the rt. As before, A stands for the total amount of money in the account, after t years have passed. P is the principal amount invested in the account. And r is the annual interest rate. Remember, e is not a variable. It's that very special number we've just discussed. What if we kept Grant's situation exactly the same as before. A principal of $20,000 and annual interest rate of would he have after five years then? Don't forget to use our new equation And please round to two decimal places.
We know that A equals P times e to the rt. We're looking for A, and here's what we know. P equals 20,000, r equals 0.04, and t equals 5. Let's plug these numbers in. That gives us 20,000 times e to the 0.2. Now, it's time to use our calculator. I'll type in 20,000, multiply that by e and take e to the power of this lesson in one go. Now I will just press Enter and we'll see what happens. We get $24,428 and if we round, 6 cents. Grant definitely makes some more money with continuously compounding interest than he did with the other options we checked out.