 This video will talk about exponential functions and applications. If we remember about exponential functions, we know that it's f of x is equal to a times b to the x, and a can't be equal to zero, b has to be greater than zero and not equal to one, and that constant b is going to be the factor that we see being multiplied over and over and over again. So there's a formula, p equal a times b to the little t over big t, that we can use to make models. p is, we'll put it in quotes, population, because it's not always a population of something, but it's an amount of same things, but it's often population, so we'll call it p. a is the original amount that we started with, b is the multiplying factor, and t, little t, is time, and big t is how often this factor actually occurs. Let's look at an example. The number of ants at a picnic is growing rapidly. At 11 a.m., five ants start the picnic. Each hour after 11 a.m., three times as many ants have found the picnic, and we want to find a of h, instead of p, we'll say a of h, represents the number of ants in the picnic, h hours after 11 a.m. So let's see what we have. a of h is going to be the number of ants. That's what they told us. That's our population. And the a, in this case, remember that's your original amount, and they told us that we said five ants had started there at 11 a.m. So we have five ants as our a. And the factor that we have going here is that each hour after 11 a.m., three times as many ants have found the picnic. So b is going to be three. Then we've got little t, and that's going to be our unknown. That's actually, in our case, going to be the h. And then the big t in our formula is how often is this happening? And it tells us right here that each hour. So that's every one hour. So if we apply it, then we have a of h is equal to our a, which is five, times our b, which is three, raised to the t, which, or h, since we're talking about hours, which we don't know what it is, and it would be divided by one. So it's really, in this case, just over, just to the h. So five times three of the h. So now, using that, again, it was a to the h, five, times the common factor of three that keeps multiplying to the t, or h. Keep your variables consistent. This is an h. This needs to be an h. So it says, estimate numerically, that is with a table, when there will be 11,000 ants at the picnic. That means we need to come in here to y equal and plug in our new equation. So five and then times three, care at x. And then we're going to go look at our table. We're looking for 11,000. So I'm just going to scroll down here and it happens very quickly. We have seven at h equal seven. We have 10,935. And at h equal eight, we have 32,805. So somewhere in the seventh hour and close to the seventh hour, it's going to be when we have this 11,000 ants. So during the seventh hour, well, let's be more specific. We'll remember that it started at 11am. That was our beginning time. So one more hour would take us to noon and then six more hours, we would have six o'clock. So 6pm somewhere between actually maybe we would be better to say between 6 and 7pm is when we're going to have 11,000 ants. So how many ants will there be at 11pm? Well, that's a time. So we need to figure out how many hours since 11am has happened. Well, if it's 11pm, that means that h is going to be equal to 12. So back to our formula. A, and now we're going to look at for h to be the 12, is equal to the original five times or three times the more we get each hour. And it's going to be to the 12. So coming back over to our calculator again, we can say five times three carat 12. And find out that that is a lot of ants. How about two, six, five, seven, two, oh, five. Two million ants plus. That is a lot of ants. Okay, let's look at another problem. This one is a half-life problem. And there's a formula for half-life that says A, the amount is equal to A, the original amount, little A, the original amount. But our base is going to be half because it's a half life, we're taking half. But then we've got this T and since it's a half-life problem, instead of using that capital T, we usually just put half-life down here. Same thing. How often does it happen? Well, the half-life would be how often having happens. So find an equation for this situation where we have a hydrogen has a half-life of 4500 days. And we want to know how many are going to be remaining with a sample of 500. So this is our little A. So we have A is equal to starting with 500 hydrogen atoms. And we have half that's going to happen every, so T divided by 4500. So here's our function just so that we remember what it is. And I put it as 0.5 because that'll be easier to put in our calculator. So it estimate the amount remaining after 50 years. And if we go back and look at our problem that we just did, it said 4500 days. So we have to figure out how many A's there are in 50 years. Well, remember there are 365 days in one year. So we have 50 years. We multiply that by 365 days per year. And we're going to get 18,250. So now we're doing H of, I called it H in earlier. I said A. We'll call it H. So H of 18,250, because it's a time, is equal to our 500 original times the half life, which we're calling 0.5, raised to the 18,250 over 4500. So if we come to our calculator, we say 500. And then in parentheses, 0.5. And then care it, but we need our fraction exponent in a parentheses. So 18,250 divided by 4500. And then close the parentheses for that exponent. And we find out that in 50 years, there are 30.07 atoms left, atoms of hydrogen left. One more example. A population starts with 500 members and doubles every 10 years. And we want to make a table to see what's happening here. So we have years and we have the population of deer. So in the first year or zero year, when we started, there were 500 deer. And it takes 10 years to double. So in 10, we're going to have 1,000, double the 500. And in 20 years, we're going to have double 1,000. So that would be 2,000. And in 30 years, we'd have 4,000. And that should be enough for us to see the pattern. Because I believe that next thing they're going to want us to do is make an equation out of this. So let's see what we know. A is 500. We started with 500 deer. And then what's happening here is we are multiplying by 2, right? It's doubling, it tells us. So that means that our B is 2. And T is going to be in years. And the big T of that formula, remember that formula was P is equal to little a times B to the little t divided by big T. And big T, how often does it happen? It happens every 10 years. So if we want to write our formula then P, or they may ask us something later, but right now we're going to use P is equal to a, which is 500, times our multiplying factor, which is doubling, or 2, to the T divided by 10. So here we have our function. How many deer will there be in 23 years? We started in years, we're still in years, so we don't have to worry about changing units here. We just have to plug and check. So we want to know what's the population in 23 years. So 500 times 2 to the 23 over 10. And again, to our calculator we go, 500 times 2 carat parentheses, 23 divided by 10. And we find out that there is 2,462. Let's just say, because I really don't want parts of deer, let's say there's approximately 2,462 deer in 23 years.