This video is provided as supplementary
material for courses taught at Howard Community
College and in this video I’m going to show how to write
growth function. The problem I have says a colony of
bacteria increases from 200 to 250 in 14 days. Write an exponential growth
function for the number of bacteria in the colony as a
formula for exponential
growth and decay, which is a A(t) equals a-sub-0 times e raised to the kt. Now let me explain what all
this means. For A(t) think of A as meaning ‘amount’ and A(t) would mean the amount to
something that you have after a certain period of time. Well
in our problem, we have 250 bacteria after 14 days. So A(t) is going to be 250. For a-sub-0…. Whenever you have a sub-0, that usually tells you that you’re looking at a number at the
beginning of a process. So this should be the amount of bacteria
at the beginning. Well, at the beginning we had 200.
Then there’s the number e and want to raise that to the kt. k is
a constant we’re going to have to look for, but we know what t is. t is 14 days. So I’ll have e^14k. I wanna find out what k is.
So let’s simplify this by dividing both sides by 200.
That will get rid of the 200 on the right side. The left side is gonna be divided by 200. I can divide by 10 first. So that’s 25 over 20 and I can turn that into 5/4. 5/4. That equals e^14k. We want to solve for k. k is an exponent, so I’ll take the logarithms of both sides and since I’ve got an e, I’ll take the
natural log. So I’ll have ln, the natural log, of 5/4 equls ln of e to the 14k. Whenever you’ve got an exponent as part the logarithm, we can make it into
a coefficient. So let’s just take that 14k and put it in front of the logarithm. Then I’ve got ln of e, which is just 1. So I’ll cross that out. Now it’s ln (5/4) equals 14k. I’ll divide both sides by 14 and that means k equals ln (5/4) over 14. I’ll use a calculator
and find out what this is. We went ln, the natural log, of 5/4, and we’ll divide that by 14. So I’ve got k equals .0159388… and this keeps going. I’m gonna round that to .0159. So now all we have to do is take that
number, .0159, and plug it back into our original formula
where the k was. Okay, so this means that our function is going to be A(t) equals A-sub-0 times e^.0159t. Notice that .0159 is positive. When you have a growth function, k is
going to be a positive number. Let’s look at a decay function now. So this says phosphorus-32, or p-32, has a half-life of 14.29 days. Write an exponential decay function for
p-32 as a function of time. So we’ll start again
with that general formula, A(t) equals A-sub-0 times e^kt. So whenever you have a half-life problem,
what that means is whatever you start with, at the end of
the half life period, you’re going to have half as much. So if we make believe we start with 1, then we could say that A(t) is going
to be 1/2, or .5. So let’s just put .5 in here. That means A-sub-0 would be 1. And then we’re going to have e^kt. We’re looking for k, but we know
what t is, it’s 14.29 days. So that’s e^14.29k. We want to find out what k is so, we’ll take the natural log of both sides. So I’ll have ln of .5 equals ln of e to the 14.29k. I’ll take the exponent and make it
a coefficient. So ln of .5 equals 14.29k times the natural log of 3, and the natural
log of e is 1. So I’ll cross that out, and now we want to divide both sides by 14.29 so the k will be by itself on the right side.
The left side will be the natural log of .5 over 14.29. And the calculator can find out
what that is. So we want the natural log of .5 divided by 14.29. So k equals -.0485057… and that keeps going too. So let’s round this to just -.0485, and we’ll put that number in where the k was. So I’ll have e to the -.0485t. So A(t) equals a-sub-0 times e to the -.0485t. And you notice that the k-value here is negative. When you
have an exponential decay, your k is going to be a negative number. Okay, that’s how you go about doing these.
Take care, I’ll see you next time.