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

exponential growth and decay functions. So let’s start with an exponential

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

function of time. So we can start with the general

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.

1:34 *letter e..not number e

Your videos are really helpful. I want you to make more videos

That was very helpful. Thank you.

in the min 4:00 , I put ln 4/5 divided by 14 in the calculater and I got -0.0159388

I don't why I got negative

help me out

what is "k" …………confussing one

You a life saver 😝

The last problem is a bit confusing, where did the .5 come from and why is the 32 not used as the A•

thanks, it was helpful.

it took me 8 minutes to understand a whole week lecture

I think the half-life is a confusing concept for many people. It is easier to understand exponential functions if "k" is regarded as an interest rate (positive) or a tax rate (negative). So the 0.0159 is the "interest rate" in the first problem, i.e. the bacteria are growing at a daily rate of 1.59% and the P-32 is decaying at 0.0485 or a rate of 4.85% every day in the second problem.

Nice video