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Exponential growth & decay word problems

How do you solve word problems involving exponential growth and decay? In this video, you will learn how to use a table and a formula to find the percentage of a radioactive substance that remains after a certain time. You will also see how a common ratio, which is the factor by which the quantity changes every time period, determines the rate of change. You will use a calculator to apply the formula and get the answers. Created by Sal Khan and CK-12 Foundation.

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Video transcript

Let's do a couple of word problems dealing with exponential growth and decay. So this first problem, suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after 6 hours? So let's make a little table here, to just imagine what's going on. And then we'll try to come up with a formula for, in general, how much is left after n hours. So let's say hours that have passed by, and percentage left. So after 0 hours, what percent is left? Well, it hasn't decayed yet, so we have 100% left. After 1 hour, what's happened? It decays at a rate of 3.5% per hour. So 3.5% is gone. Or another way to think about it is 0.965. Remember, if you take 1 minus 3.5%, or if you take 100% minus 3.5%-- this is how much we're losing every hour-- that equals 96.5%. So each hour we're going to have 96.5% of the previous hour. So in hour 1, we're going to have 96.5% of hour 0, or 0.965 times 100, times hour 0. Now, what happens in hour 2? Well, we're going to have 96.5% of the previous hour. We will have lost 3.5%, which means that we have 96.5% of the previous hour. So it'll be 0.965 times this, times 0.965 times 100. I think you see where this is going, in general. So in the first hour, we have 0.965 to the first power, times 100. In the zeroth hour, we have 0.965 to the zeroth power. We don't see it, but there's a 1 there, times 100. In the second hour, 0.965 to the second power, times 100. So in general, in the nth hour-- let me do this in a nice bold color-- in the nth hour, we're going to have 0.965 to the nth power, times 100 left of our radioactive substance. And oftentimes you'll see it written this way. You have your initial amount times your common ratio, 0.965 to the nth power. This is how much you're going to have left after n hours. Well, now we can answer the question. After 6 hours how much are we going to have left? Well, we're going to have 100 times 0.965 to the sixth power left. And we could use a calculator to figure out what that is. Let's use our trusty calculator. So we have 100 times 0.965 to the sixth power, which is equal to 80.75. This is all in percentages. So it's 80.75% of our original substance. Let's do another one of these. So we have, Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is-- oh actually, there's a typo here, it should be 8%-- the rate of increase is 8% annually, how many stores does the restaurant operate in 2007? So let's think about the same thing. So let's say years after 1999. And let's talk about how many stores Nadia is operating, her fast food chain. So 1999 itself is 0 years after 1999. And she is operating 200 stores. Then in 2000, which is 1 year after 1999, how many is she going to be operating? Well, she grows at the rate of 8% annually. So she'll be operating all the stores that she had before plus 8% of the store she had before. So 1.08 times the number of stores she had before. And you're going to see, the common ratio here is 1.08. If you're growing by 8%, that's equivalent to multiplying by 1.08. Let me make that clear. 200 plus 0.08, times 200. Well, this is just 1 times 200 plus 0.08, times 200. That's 1.08 times 200. Then in 2001, what's going on? This is now 2 years after 1999, and you're going to grow 8% from this number. You're going to multiply 1.08 times that number, times 1.08 times 200. I think you get the general gist. If, after n years after 1999, it's going to be 1.08-- let me write it this way. It's going to be 200 times 1.08 to the nth power. After 2 years, 1.08 squared. 1 year, 1.08 to the first power. 0 years, this is the same thing as a 1 times 200, which is 1.08 to the zeroth power. So they're asking us, how many stores does the restaurant operate in 2007? Well, 2007 is 8 years after 1999. So here n is equal to 8. So let us substitute n is equal to 8. The answer to our question will be 200 times 1.08 to the eighth power. Let's get our calculator out and calculate it. So we want to figure out 200 times 1.08 to the eighth power. She's going to be operating 370 restaurants, and she'll be in the process of opening a few more. So if we round it down, she's going to be operating 370 restaurants. So 8% growth might not look like something that's so fast or that exciting. But in under a decade, in only 8 years, she would have gotten her restaurant chain from 200 to 370 restaurants. So over 8 years, you see that the compounding growth by 8% actually ends up being quite dramatic.