# Video: Introduction to Exponential Growth and Decay Story Problems

Applying our knowledge of exponential growth and decay and the general format of exponential functions to problems requiring us to form and solve exponential equations.

09:49

### Video Transcript

In this video we will be learning how to apply our knowledge of geometric sequences and exponential growth and decay to wordy or story problems. We will be looking at how to write exponential growth and decay formulae and how to use them to solve problems.

Looking now first question then, the population of a town is increasing at a rate of two point six percent per year. Assuming this rate continues and that the current population is twenty-six thousand, seven hundred and fifty: a) write a formula to calculate the population 𝑝 in 𝑥 years time and b) use the formula to find the population nine years from now.

Okay so let’s just write down the definition of our variables first. So 𝑥 is the time in years from now. So when 𝑥 is zero, we’re talking about now. And 𝑝 represents the size of the population. Now the general format of an exponential growth formula is of this form. So 𝑝 is equal to some initial amount times the multiplier to the power of 𝑥, in this case the number of years.

And the starting population we were talking in the question is twenty-six thousand, seven hundred and fifty. And to work out the multiplier, remember we’re adding two point six percent of the population every year. So if we start up with a hundred percent of the population and we add two point six percent to that, we’ve got a hundred and two point six percent. Now to work out a hundred and two point six percent of something, we need to think that a hundred and two point six percent means a hundred and two point six per a hundred out of a hundred divided by a hundred. And a hundred and two point six divided by a hundred is one point o two six.

So we can put these values into a formula. So the formula becomes 𝑝 equals twenty-six thousand, seven hundred and fifty times one point o two six to the power of 𝑥. I can tidy that up slightly, but it’s still the basic formula. And that is the answer to part a. Now I’ve got to use that formula to find the population nine years from now, so when 𝑥 is equal to nine.

So this is the calculation we’re going to do 𝑝 is equal to twenty-six thousand, seven hundred and fifty times one point o two six to the power of nine. And my calculator gives me this answer here: thirty-three thousand, seven hundred and one point five six two two six and so on. Now unless we’re trying to interpret all the decimal places as a measure of the completeness of the latest birth of a baby, I think it probably doesn’t make much sense to use so many decimal places in this. When we’re talking about population, we just talk about rounding this to the nearest whole number. So it gives us thirty-three thousand, seven hundred and two people. Now there is a case for saying because we don’t have a whole extra person, we should round that number down to thirty-three thousand, seven hundred and one. And that’s a fine argument; I’m not gonna complain about that. But for just purposes of this illustration, we have rounded to the nearest whole number.

An outbreak of ant lurgy disease affects a large colony of ants. Fifty ants are infected to start with, and each day the number of infected ants trebles. Write an expression for 𝐴 𝑑 the number of ants with the disease at the end of day 𝑑. Use it to calculate how many ants will be infected after two weeks.

So a general format for these exponential formulae is the amount is equal to the initial amount times the multiplier to the power of the time, in this case the number of days. And we’re told in the questions that fifty ants are infected to start with. And that each day the number of infected ants trebles. So the multiplier is times three. So let’s plug those numbers into our formula.

So the number of ants infected is fifty times three to the power of 𝑑. But let’s just write that a little bit more neatly. Well that’s the first bit done. Now we’re gonna use the formula to calculate how many ants will be infected after two weeks. Well our units are in days and we’ve been asked a question in weeks. So the first thing I’m gonna do is convert that into days. And two weeks is two lots of seven; that’s fourteen days. So we’re gonna put a value for 𝑑 of fourteen in our formula. So the amount of ants that have the disease on day fourteen is fifty times three to the power of fourteen. And that gives us an answer of two hundred and thirty-nine million, one hundred and forty-eight thousand, four hundred and fifty ants. Now it’s possible that’s more than the entire population of this ant colony. So it’s worth noting that if that’s the case, then the model would break down.

For our next question, the function 𝑉 𝑡 is equal to 𝐴 times 𝑚 to the power of 𝑡 represents the amount of money in a savings account after 𝑡 years. 𝐴 represents the initial investment, and the annual interest rate is six point five percent. After the initial investment is made, interest is added at the end of each year to the account, and no other deposits or withdrawals are made. Find the value of 𝑚.

Now we know that this is an exponential growth question because we’ve told the general format here, and we’re given some information here about no extra money being paid in or money been taken out apart from the interest being added every year. now this is a slightly unusual question because we’re not told the actual amount that was saved. And we’ve got to find out the value of one of the parameters in our formula. So let’s start off by just writing out the general format of an exponential growth formula. And that would be the amount of money — the value of the account if you like — is equal to the initial amount times the multiplier to the power of the amount of time, the number of years.

Then we’re told that 𝐴 represents the initial investment even that we don’t have an actual number for that. And we’re told that we are adding six point five percent each year. So we’re starting with a hundred percent and then we’re adding six point five percent. So after one year we got a hundred and six point five percent of our initial amount and a hundred and six point five percent means out of a hundred. And a hundred and six point five over a hundred means a hundred and six point five divided by a hundred. That gives us one point o six five. So to work out a hundred and six point five percent of something, we use a multiplier of one point o six five.

Now we can put those values into a general formula: 𝐴 is the initial amount and one point zero six five is the multiplier. So the value of 𝑚 is one point o six five. So this question really just involved using a knowledge of the format of the general equation for exponential growth or decay and then just checking off the corresponding values with the equation that they gave us in the question.

One last question then, a car loses twelve percent of its value each year. Write an expression for 𝑉 𝑡, the value of the car after 𝑡 years, given that it cost thirty thousand dollars when new. Use the expression to calculate the value of the car after twenty-two years. So first thing let’s write down a general formula for exponential growth or decay.

Well the value is equal to the initial amount times a multiplier to the power of 𝑡, where in this case 𝑡 represents the number of years. The question tells us that the car was worth thirty thousand dollars when new. And we have to decrease the value of the car by twelve percentage. So we’re gonna come up with a multiplier that decreases the value by twelve percent.

Well if we started off with a hundred percent of a value and we take away twelve percent of that value, that would leave us with eighty-eight percent of the value. So we’ve got to have a multiplier that calculates eighty-eight percent of the previous amount. Now eighty-eight percent means eighty-eight out of a hundred. And eighty-eight over a hundred or eighty-eight divided by a hundred is nought point eight eight. Now if I use nought point eight eight as a multiplier — if I multiply a number by nought point eight eight — I’m effectively calculating eighty-eight percent of it; well that’s the same as reducing it by twelve percent. So that is gonna be my multiplier.

So let’s pop those values into a formula. And then we can just write it out a bit more neatly to give our answer to the first bit. So the expression is thirty thousand times nought point eight eight to the power of 𝑡.

Now we got to use the expression to calculate the value of the car after twenty-two years. So that basically means that we’re gonna put 𝑡 is equal to twenty-two into a formula to work out the value after twenty-two years. So if we replace 𝑡 in a formula with twenty-two and then round our answer to two decimal places because it’s money, we get the car is worth one thousand, eight hundred and one dollars and ninety-four cents.

So in summary all of these questions are really about using a formula for exponential growth or decay: 𝑦 is equal to the initial amount times the multiplier to the power of 𝑥. And in the story problems, you’re often given the initial amount in the question. And the multiplier we have to work out. First what we think is it a percentage increase or is it a percentage decrease? And then we’ll take a hundred percent and leave that at the increase on in percentage or will take the decrease off in percentage. And then we’ll express that percentage over a hundred to work out a multiplier, which we can then just put it into the formula.