Adding versus multiplying
Tonight's idea
Two columns, A and B. In both, \(x\) goes up by 1 each row.
| \(x\) | A | B |
|---|---|---|
| 0 | 5 | 1 |
| 1 | 8 | 2 |
| 2 | 11 | 4 |
| 3 | 14 | 8 |
| 4 | 17 | 16 |
| 5 | 20 | 32 |
A has a constant difference, so it is linear: \(y = 3x + 5\).
B has a constant ratio. Divide each value by the one before: \(\tfrac21\), \(\tfrac42\), \(\tfrac84\), always 2. A function like this is exponential.
\(y = a \cdot b^x\). Here \(a\) is the value at \(x = 0\) and \(b\) is the multiplier for each step.
Why this rule? Start at \(a\). Each step multiplies by \(b\), so after \(x\) steps you have multiplied by \(b\) a total of \(x\) times: \(a \cdot b^x\). At \(x = 0\) there are no steps, and \(b^0 = 1\) leaves just \(a\). Column B is \(y = 1 \cdot 2^x\).
If \(b\) is bigger than 1, \(y\) grows. If \(b\) is between 0 and 1, \(y\) shrinks: 80, 40, 20, 10 has \(b = \tfrac12\).
Adding gives the same gain every step. Multiplying gives a gain that itself grows, which is why B wins in the end.
Watch one: 5, 15, 45, 135
These are the values of \(y\) at \(x = 0, 1, 2, 3\). Differences: 10, 30, 90. Not constant, so not linear.
Ratios: \(\tfrac{15}{5} = 3\), \(\tfrac{45}{15} = 3\), \(\tfrac{135}{45} = 3\). Constant, so exponential with \(b = 3\). The value at \(x = 0\) is 5, so \(a = 5\).
The rule is \(y = 5 \cdot 3^x\). Check \(x = 2\): \(5 \times 9 = 45\). Predict \(x = 5\): \(5 \times 243 = 1215\).
Race a line \(a \times n\), the square \(n^2\) and a multiplier \(b^n\) on one chart. Slide \(a\) up and \(b\) down, and read off the step where the multiplier overtakes for good: mathdetours.com/demos/growth.html.
Your turn
- Each list gives \(y\) at \(x = 0, 1, 2, 3\). Linear, exponential or neither? Give the constant difference or ratio. (a) 7, 11, 15, 19. (b) 3, 6, 12, 24. (c) 1, 4, 9, 16. (d) 80, 40, 20, 10.
- (a) Write the rule \(y = a \cdot b^x\) for lists (b) and (d) in question 1, and find \(y\) at \(x = 6\) for each. (b) Another exponential table has \(y = 12, 36, 108\) at \(x = 1, 2, 3\). Find \(b\), then \(a\). Careful: \(a\) is not 12.
- Growing by 5% means multiplying by 1.05. (a) A town of 20,000 people grows by 5% a year. Write a rule for its population after \(x\) years, and find it after 3 years. (b) A car bought for $15,000 loses 20% of its value each year. What is \(b\)? What is the car worth after 3 years?
- An exponential function passes through \((2, 18)\) and \((4, 162)\). (a) Two steps multiplied \(y\) by how much? Find \(b\) (it is positive) and \(a\). (b) Find the linear function through the same two points. (c) Which function is larger at \(x = 3\)? At \(x = 6\)?
Stretch: differences of an exponential
The differences of 2, 6, 18, 54, 162 are 4, 12, 36, 108: exponential again, same ratio. Show that the differences of any \(a \cdot b^x\) are exponential with the same \(b\). For which \(b\) are the differences the original values?
Answers
- (a) Linear, difference 4. (b) Exponential, ratio 2. (c) Neither: differences 3, 5, 7 and ratios 4, 2.25, 1.78 both change. (d) Exponential, ratio \(\tfrac12\).
- (a) \(y = 3 \cdot 2^x\), and \(3 \times 64 = 192\). \(y = 80 \cdot \left(\tfrac12\right)^x\), and \(80/64 = 1.25\). (b) \(b = 36/12 = 3\). Step back once from \(x = 1\): \(a = 12/3 = 4\), so \(y = 4 \cdot 3^x\).
- (a) \(y = 20{,}000 \times 1.05^x\). After 3 years: \(20{,}000 \times 1.157625 = 23{,}152.5\), about 23,150 people. (b) Losing 20% leaves 80%, so \(b = 0.8\). \(15{,}000 \times 0.8^3 = 15{,}000 \times 0.512 = \$7{,}680\).
- (a) \(162/18 = 9\), and two steps multiply by \(b^2\), so \(b = 3\). Step back twice from 18: \(a = 18/9 = 2\). \(y = 2 \cdot 3^x\). (b) Slope \(\dfrac{162 - 18}{4 - 2} = 72\), so \(y = 72x - 126\). (c) At \(x = 3\): exponential 54, line 90. At \(x = 6\): exponential 1458, line 306. The line is ahead between the two shared points and the exponential is ahead after them.
Stretch. One step takes \(a \cdot b^x\) to \(a \cdot b^{x+1} = a \cdot b^x \cdot b\). The difference is \(a \cdot b^x \cdot b - a \cdot b^x = a(b - 1) \cdot b^x\): exponential, starting value \(a(b - 1)\), same multiplier \(b\). It equals the original when \(b - 1 = 1\), so \(b = 2\). Doubling is the growth whose gain each step is what you already have.