00:01
All right, so we have some information here about the amount of deer in two different years, and we want to create an equation where t is years since 1970, and we just have to figure out a and b for this specific equation.
00:14
So this means that in 1970, t equals zero, and nt, the number of deer, is that 20 ,000.
00:22
And then in 1990, 20 years have passed, so that means t is 20, and then the n of t is this 120 ,000.
00:32
So first thing i am going to do is i'm going to take this first set of information, and i'm going to plug it in here.
00:39
The n of t is that 20 ,000, the amount of deer.
00:43
We don't know a, we don't know b, but we do know that t is zero.
00:48
The reason that this is really helpful to us is that anything to the zero power is just one, so b to the zero is one.
00:57
So we end up with 20 ,000 equals a times one, so 20 ,000 is our a value.
01:05
Now what i'm going to do is i'm going to kind of go back to my original equation and do the same things, plug in the information we have from that kind of second time.
01:15
So the n of t there is 120 ,000 equals a times b to the t, and now we know a, a is 20 ,000.
01:25
We still don't know b, but in this specific situation where we're using this 120 ,000, our t is 20.
01:33
So i'm plugging in the a from over here, and i'm plugging in the 20 and 120 ,000 from up there.
01:45
So here we're just going to try to solve for b.
01:48
The first thing i need to do is get rid of that 20 ,000 that's multiplying in front of it, so i'm going to divide that on both sides.
01:54
120 ,000 divided by 20 ,000 gives us 6 equals b to the 20th.
02:13
Now we want to get rid of that power of 20.
02:16
Anytime you're getting rid of an exponent, it's like undoing a squared, so it'd be the same thing as undoing x squared.
02:24
You want to do the matching root, so this is a power of 20, which means we're taking the 20th root, and we want to do that to both sides.
02:32
That'll get rid of that power of 20 and just leave us with a b...