Lesson Video: Applications of Exponential Functions Mathematics • 9th Grade

In this video, we will learn how to solve real-world problems involving exponential functions.


Video Transcript

In this video, we will learn how to solve real-world problems involving exponential functions. To do that, we’ll remember the general exponential function forms. We’ll start with 𝑓 of π‘₯ equals 𝑏 to the π‘₯ power, where the base 𝑏 is a positive number other than one. However, when we want to model real-world data in an exponential form, we need to modify this slightly. We use 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power. It’s still true that 𝑏 is a positive number other than one. The independent variable, the power, usually represents some unit of time. And the variable 𝐴 represents the initial value of what the function is measuring.

We can easily see what the initial value is when π‘₯ is zero, when no time has passed. We’ve already said that the base 𝑏 must be a positive number other than one. This value for the base 𝑏 tells us something about the rate at which the quantity changes over time. It tells us how our initial value is changing. And these changes fall in one of two categories. We’ll either have a function that increases over time or a function that decreases over time. An increase represents exponential growth, while a decrease represents exponential decay. When we’re dealing with exponential growth, the 𝑏-value is greater than one. And when we’re dealing with exponential decay, the 𝑏-value will be less than one.

But remember, we’ve already said that the 𝑏-value must be positive. And that means for exponential decay the 𝑏-value must fall between zero and one. Before we look at any examples, there’s one more thing we need to note about this rate, this 𝑏-value. In exponential modeling, very often we’ll be dealing with percent increase or decrease. And we need to think carefully about how we translate a percent increase or a percent decrease into a base exponential value.

Let’s say we want to model a three percent decrease over time. Since we know that this is a decrease, we know that we’re looking for a base exponential value between zero and one. We also know that three percent is three out of 100 or, written as a decimal, 0.03. However, this function models the change in the initial quantity over time. And that means we’re not modeling how much we lost; we’re modeling how much is remaining over every unit of time. And if three percent is the decrease, the 97 percent remains. And that means as a function, we could call this 𝑓 of π‘₯ equals π‘Ž times 0.97 to the π‘₯ power. It’s also possible to model this as the fraction 97 over 100 to the π‘₯ power, though it’s more common in this type of modeling to use the decimal form.

Now let’s think about how we would model a three percent increase. That percentage written as a decimal is 0.03. This is telling us that every unit of time, we’re gaining three percent of the initial amount. And when we’re working with exponential increase, that 𝑏-value will be greater than one. And what’s happening here is we have 100 percent of what we started with plus the three percent increase every unit of time. And we model that as 𝑓 of π‘₯ equals π‘Ž times 1.03 to the π‘₯ power. In modeling, there are a few common ratios we see again and again. If the value is doubling every unit of time, then the 𝑏-value equals two. And if the changes over time is half, the 𝑏-value is equal to one-half. Now we’re ready to consider some examples.

The number of people visiting a museum is decreasing by three percent a year. This year there were 50,000 visitors. Assuming the decline continues, write an equation that can be used to find 𝑉, the number of visitors there will be in 𝑑 years’ time.

When we’re dealing with the percent decrease, we’re not dealing with a linear function, so we know we’ll need an exponential function to model this. This means we’ll use the general form 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power. Our 𝑏-variable is the rate of change. The π‘₯-variable represents the unit of time we’re measuring. And 𝐴 represents the initial value. We need to be clear here that we want to model the number of visitors there will be at the museum.

A decrease in three percent of the visitors means that 97 percent of the visitors are maintained. Since we’re modeling the number of visitors, we’ll use 97 percent. We’ll write this percent in decimal form 0.97. We know that our π‘₯-value is being measured in 𝑑 years. And our starting value, our initial value, are the 50,000 visitors from this year. Our equation is modeled with a capital 𝑉 so that we have the equation 𝑉 equals 50,000 times 0.97 to the 𝑑 power.

In our next example, we’re given a model, and we need to interpret data from that model.

A population of bacteria decreases as a result of a chemical treatment. The population 𝑑 hours after the treatment was applied can be modeled by the function 𝑃 of 𝑑, where 𝑃 of 𝑑 equals 6000 times 0.4 to the 𝑑 power. What was the population when the chemical was first applied? And what is the rate of population decrease?

Our function, that’s the general form 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power. In this form, the 𝐴 represents the initial value, which means we identify 6000 as the initial value. One way to check that this is true would be to plug in 𝑑 equals zero. When zero time has passed, we know that the bacteria has its original population. 0.4 to the zero power equals one, and 6000 times one equals 6000, which confirms the value for the initial population. So we turn our attention to the rate of population decrease.

For this model, we know that the 𝑏-value tells us something about the rate. The function is written so that it tells us how many of the bacteria remain after 𝑑 hours. If 0.4 remain, then 0.6 is the amount that has been decreased. If we start with one whole amount, some rate of decrease will give us 0.4 of the population that remain. And that value here is 0.6. We usually want to write this in percent form, so we say that there was a 60 percent decrease. This model is showing us that after the chemical treatment, there’s a 60 percent decrease per hour of the population of bacteria.

In our next example, we’ll write a model and then use it to solve for a quantity after a certain amount of time.

A microorganism reproduces by binary fission, where every hour each cell divides into two cells. Given that there were 15,141 cells to begin with, determine how many cells there were after five hours.

Since this microorganism is reproducing, we will expect more cells and not less, which means we’re expecting exponential growth. Our unit of time is every hour. That means we can let 𝑑 be equal to the hours after the initial count. If every hour one cell divides into two cells, one cell becomes two cells in an hour. After an additional hour, the two cells become four. This represents a doubling of the cells every hour.

So we’ll need to take our exponential form 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power, where 𝐴 is our initial value, 15,141. 𝑏 is the rate. Since our rate is doubling, 𝑏 is equal to two. And our variable will be 𝑑. It’ll be units of time. We now wanna take this function and use it to solve for how many cells there were after five hours, which means we need to calculate 15,141 times two to the fifth power, which is 484,512. After five hours, we can expect that this microorganism will have 484,512 cells.

In our final example, we’ll take some data we’re given and use it to create a population model.

The US census is taken every 10 years. The population of Texas was 3.05 million in 1900 and 20.9 million in 2000. By modeling the growth as exponential, answer the following questions. Write an exponential function in the form 𝑃 of 𝑑 equals 𝑃 nought times π‘˜ to the 𝑑 power to model the population of Texas, in millions, 𝑑 decades after 1900. Round your value of π‘˜ to three decimal places if necessary. According to the model, what was the population of Texas in 1950? Give your answer to three significant figures. And finally, rewrite your function in the form 𝑃 of 𝑦 is equal to 𝑃 nought times 𝑏 to the 𝑦 power, where 𝑦 is the time in years after 1900. Round your value of 𝑏 to four decimal places.

Let’s start with what we know. In 1900, the population was 3.05 million. And by 2000, that value was 20.9 million. We know the general form of the exponential function 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power. We’re following this general form with the function 𝑃 of 𝑑 equals 𝑃 nought times π‘˜ to the 𝑑 power, where our 𝑑 represents time in decades after 1900. And that means our initial value 𝑃 nought should be the population in 1900. Since we’re working in millions, we can leave this as 3.05. π‘˜ is our unknown value. To solve for π‘˜, we can use our other data point.

We know that in 2000 there was a population of 20.9 million. We also know that 2000 is 10 decades after 1900. Plugging all of this in, we can use that information to solve for π‘˜. To get π‘˜ by itself, we divide both sides of the equation by 3.05, which gives us 6.8524 continuing is equal to π‘˜ to the 10th power. Instead of rounding this, we’ll just leave it in our calculator as is. To isolate π‘˜, we’ll raise both sides of this equation to the one-tenth power. π‘˜ to the 10th power to the one-tenth power equals π‘˜, and 6.8524 continuing to the one-tenth power equals 1.212228 continuing. We want to round this π‘˜-value to three decimal places.

The fourth decimal place has a two. So we say that π‘˜ equals 1.212. This π‘˜-value that’s greater than one tells us we’re dealing with population growth. And if we think of the decimal 0.212 as a percent, we can say that the population is growing at a rate of about 21.2 percent every decade. And we’ve created a model to calculate what the population would be 𝑑 decades after 1900. 𝑃 of 𝑑 equals 3.05 times 1.212 to the 𝑑 power. Using this model, we want to estimate what the population was in 1950. 1950 is 50 years after 1900, which is five decades. To calculate this, we wanna take 𝑃 of five, 3.05 times 1.212 to the fifth power, which is equal to 7.9765 continuing. Three significant figures in this case would be to the second decimal place. If we round to the second decimal place, we get 7.98. Based on our model, we can expect that the population of Texas in 1950 would have been 7.98 million.

For part three of this question, we want to rewrite our exponential model, where our unit of time is years instead of decades. It will be a really similar process to what we did in the first part. We’ll still have the same initial value of 3.05. But to solve for 𝑦, we’ll use our second data point. In 2000, the population was 20.9 million. And that was 100 years after our initial value. So we’ll plug in 100 for 𝑦, and then we can solve for 𝑏. By dividing both sides of the equation by 3.05, we get 6.8524 continuing equals 𝑏 to the one hundredth power. Again, we don’t wanna round this 6.852 continuing yet. We’ll just leave it in our calculator so that we can raise both sides of this equation to the one over 100 power.

𝑏 to the one hundredth power to the one over 100 power equals 𝑏. And 6.8524 continuing to the one over 100 power equals 1.01943 continuing. We’re rounding to four decimal places this time, which means we’ll use 𝑏 equals 1.0194. Our 𝑏-value is less than our π‘˜-value. According to the 𝑏-value, the population growth per year was 1.94 percent versus a 21.2 percent population growth decade over decade. So for our yearly model, we have 𝑃 of 𝑦 equals 3.05 times 1.0194 to the 𝑦 power.

From here, we’re ready to review our key points. The exponential form for modeling real-world problems is 𝑓 of π‘₯ equals 𝐴 times 𝑏 to the π‘₯ power, where π‘Ž is the initial quantity, 𝑏 is how the quantity changes over time β€” when 𝑏 is greater than one, we have exponential growth, when 𝑏 is between zero and one, we have exponential decay β€” and π‘₯ is the unit of time.

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