Learning Objectives
- Apply and construct exponential models using the form y = a(b)x
Just as in linear growth, we can model quantities that grow in a non-linear pattern. Consider exponential growth.
Exponential Models
Consider the older brother who offered to pay his younger brother 1 penny per day to help him mow lawns and rake leaves. The younger brother said, “No thanks! But, I will work for you for 30 days in a row. You will pay me 1 penny on the first day and then each day after that you will double that amount you paid me the previous day.”
Let’s model that situation with a table and a graph. Let x represent the number of days, with zero representing Day 1. Let y = the amount of money the younger brother earns in this model.
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Notice the table of x- and y-values, along with the graph. Younger brother would only be making $10.24 on Day 10. But, then something happens around Day 25. The pay drastically increases. By Day 30, he would be making over $5 million.
Graphs shaped like this are modeling exponential growth. Let’s examine the table to determine the equation that models the younger brother’s pay.
| Day | x = number of days (Day 1 is 0) | Pattern | y = amount of money |
| 1 | 0 | Pay on Day 1 = $0.01 | $0.01 |
| 2 | 1 | Pay on Day 2 = $0.01 x 2 (When something doubles, that means multiply by 2.) | $0.02 |
| 3 | 2 | Pay on Day 3 = $0.01 x 2 x 2 | $0.04 |
| 4 | 3 | Pay on Day 4 = $0.01 x 23 (Switch notation to using exponents.) | $0.08 |
| 5 | 4 | Pay on Day 5 = $0.01 x 24 | $0.16 |
| 10 | 9 | Pay on Day 10 = $0.01 x 29 | $10.24 |
| 25 | 24 | Pay on Day 25 = $0.01 x 224 | $167,772.16 |
| 26 | 25 | Pay on Day 26 = $0.01 x 225 | $333,544 |
| 27 | 26 | Pay on Day 27 = $0.01 x 226 | $671,088.64 |
| 28 | 27 | Pay on Day 28 = $0.01 x 227 | $1,342,177.28 |
| 29 | 28 | Pay on Day 29 = $0.01 x 228 | $2,684,354.56 |
| 30 | 29 | Pay on Day 30 = $0.01 x 229 | $5,368,709.12 |
Letting x be the number of days after the first day, then Day 1 is 0 and Day 2 is 1, and so on. Then letting y be the amount of pay younger brother is going to receive, the last day of the month younger brother receives $0.01 x 229. Now, letting x represent any day, the equation becomes: y = $0.01(2)x .
Suppose that the younger brother said the pay would be triple the amount paid on the previous day. The model would be:
y = $0.01(3)x
If the younger brother had used the word quadruple, the model would be:
y = $0.01(4)x
Investing Money
Consider the scenario of investing money.
Suppose $5,000 is invested into an investment that earns 20% simple interest per year. To compute the interest earned in the first year, use the simple interest formula: Interest = Principle x Rate x Time (in years), or I = PRT.
After one year, the interest earned would be:
I = PRT
= ($5,000) x (0.20) x (1 year)
= $1,000
Then, adding the interest earned in the first year ($1,000) to the original investment ($5,000), there would be $6,000 in the investment after one year. This process can be written as:
Accumulated Amount = Initial Investment + Interest Earned in One Year
A = P + PRT
Using algebra, we can factor out the P to get: A = P(1 + RT)
This shows us that we can also determine the amount of accumulated after one year by multiplying P by (1 + RT).
A = $5,000 (1 + (0.20)(1 year))
A = $5,000 (1.20)
Converting 1.20 into a percent would give us 120%. Think of 120% as 100% + 20%, or the original plus the 20% increase. So, 1.20 represents a 20% increase per year.
The investment starts the second year with $6,000. We can use the same process to get the amount accumulated after the second year:
A = $6,000 (1.20)
A = $7,200 (this is the amount accumulated after 2 years)
Setting up the pattern for multiple years, we have:
| t = number of years | Pattern | y = Amount Accumulated |
| 0 | $5,000 (this is the initial investment) | $5,000 |
| 1 | $5,000 x (1.20) | $6,000 |
| 2 | $5,000 x (1.20)(1.20) (this is the accumulated amount at the end of the previous year multiplied by 1.20) | $7,200 |
| 3 | $5,000 x (1.20)(1.20)(1.20) | $8,640 |
| 4 | $5,000 x (1.20)(1.20)(1.20)(1.20) | $10,368 |
| 5 | $5,000 x (1.20)5 (switch to notation using exponents) | $12,441.60 |
The general pattern is: y = $5,000(1.20)t
Key Takeaways
The general equation for exponential growth is:
y = a(b)t
- Where a is the initial amount
- b is (1 + growth rate) or (1 + r).
- b is called the base.
- b is sometimes called growth factor.
Example
| Initial Investment | Simple Interest Rate | Model, where y = amount accumulated and t = number of years |
| $300 | 5% | y = 300(1.05)t |
| $1,000 | 8.5% | y = 1000(1.085)t |
| $25,000 | 3% | y = 25000(1.03)t |
The exponential growth model (equation) can be made more generic by letting the initial investment (principle) be some other quantity.
TRY IT
Fill in the missing cells based on the example in the first row.
| Initial Amount | Growth Rate | b = base (or growth factor) | y = amount accumulated, t = time unit |
| 30 | 1% | 1.01 | y = 30 (1.01)t |
| 150 |
Show Answer
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1.35 | y = 150 (1.35)t |
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Show Answer
|
99% |
Show Answer
|
y = 450 (1.99)t |
|
Show Answer
|
100% |
Show Answer
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y = 5 (2)t |
| 900 |
Show Answer
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3 | y = 900 (3)t |
| 6 | 300% |
Show Answer
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y = 6 (4)t |
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