Math 1 · Unit 8 · Lesson 5 · part 4 of 5
Using the models: what if, and how long?
A model is useful because you can change a number and see what happens.
Using the models: what if, and how long?
A model is useful because you can change a number and see what happens. A city has 8,000 people and grows 4% a year: 8,000(1.04)t. The 8,000 is the population now. The 1.04 is 1 + 0.04, growing 4% each year.
Change the start: 12,000 instead of 8,000
Same factor, bigger start. 12,000 is 1.5 times 8,000, so every value is 1.5 times bigger. After 10 years: 17,763 instead of 11,842. The curve has the same shape, stretched taller.
Change the rate: 6% instead of 4%
Same start, bigger factor, 1.06. The two curves begin together, and the gap widens every year. After 10 years: 14,327 against 11,842. After 20 years: 25,657 against 17,529.
Try it. The dashed gray curve stays where you started, so you can see the old and new curves together.
Goal questions ask how long. How many years until $500 at 8% compounded annually doubles to $1,000? There is no solving tool for the exponent yet, so build a table and watch for the goal, the same backward method as Unit 7, Lesson 6.
| year | 500(1.08)t |
|---|---|
| 0 | $500.00 |
| 1 | $540.00 |
| 2 | $583.20 |
| 5 | $734.66 |
| 8 | $925.47 |
| 9 | $999.50 |
| 10 | $1,079.46 |
After 9 years the balance is $999.50, fifty cents short of double. It passes $1,000 during year 10. So the doubling time is about 9 years.