6.4 Exponential Growth and Decay: ab^x and a(1 + r)^x
Key Takeaways
- In f(x) = ab^x, a is the initial value at x = 0 and b is the growth factor; b > 1 gives growth and 0 < b < 1 gives decay.
- The form a(1 + r)^x uses a rate: r > 0 gives growth, and decay is written a(1 − r)^x, so a 15% decline uses a factor of 0.85.
- Convert between forms by matching b to 1 + r, so a factor of 1.07 is a 7% increase and a factor of 0.92 is an 8% decrease.
- Exponential graphs have a horizontal asymptote at y = 0 and never reach zero, which distinguishes decay from a linear decrease that crosses the axis.
- Compound interest with n periods per year is A = P(1 + r/n)^(nt), which divides the rate and multiplies the exponent.
6.4 Exponential Growth and Decay
Competency 2 skill 9 is written with unusual precision: it names two specific forms, ab^x and a(1 + r)^x, and three contexts, depreciation, compound interest, and population growth. Expect items that hand you a scenario and ask which equation or which graph matches it.
The two forms
+---------------------------------------------------------------------------+
| FACTOR FORM f(x) = a * b^x |
| a = initial value (the y-intercept, value at x = 0) |
| b = growth FACTOR per period |
| b > 1 -> growth b = 1.25 means +25% each period |
| 0 < b < 1 -> decay b = 0.75 means -25% each period |
| |
| RATE FORM f(x) = a * (1 + r)^x growth at rate r |
| f(x) = a * (1 - r)^x decay at rate r |
| r expressed as a DECIMAL: 8% -> r = 0.08 |
+---------------------------------------------------------------------------+
The two forms describe the same functions; b = 1 + r for growth and b = 1 − r for decay. Translating between them is a routine item:
| Factor b | Interpretation |
|---|---|
| 1.06 | 6% growth per period |
| 1.5 | 50% growth per period |
| 2 | doubling, 100% growth |
| 0.94 | 6% decay per period |
| 0.85 | 15% decay per period |
| 0.5 | halving, 50% decay |
The reliable trap: reading b = 0.85 as "85% decay." It is a 15% decay, because the quantity retains 85%. Similarly b = 1.15 is 15% growth, not 115% growth.
Building a model from a scenario
Population growth. A town of 24,000 grows 3% per year.
P(t) = 24,000(1.03)^t. After 10 years: 24,000(1.03)¹⁰ = 24,000(1.3439) ≈ 32,254 people.
Depreciation. A $38,000 truck loses 18% of its value each year.
V(t) = 38,000(0.82)^t. After 4 years: 38,000(0.82)⁴ = 38,000(0.45212) ≈ $17,181.
Half-life decay. A 200 mg sample halves every 6 hours.
A(t) = 200(0.5)^(t/6), where dividing t by the half-life counts how many halvings have occurred. After 18 hours: 200(0.5)³ = 25 mg.
Compound interest. With n compounding periods per year,
A = P(1 + r/n)^(nt)
$5,000 at 4% compounded quarterly for 3 years: A = 5,000(1 + 0.04/4)^(4·3) = 5,000(1.01)¹² = 5,000(1.12683) ≈ $5,634.13.
Note both adjustments: the rate is divided by n and the exponent is multiplied by n. Doing only one is the standard error, and both wrong values appear as options.
Exponential versus linear change
This distinction is the conceptual heart of the skill.
| Linear | Exponential | |
|---|---|---|
| Changes by | a constant amount | a constant percent/factor |
| Table signature | constant differences | constant ratios |
| Form | y = mx + b | y = ab^x |
| Long-run decrease | crosses zero, goes negative | approaches zero, never reaches it |
A $30,000 car losing $3,000 per year is linear: V = 30,000 − 3,000t, and it is worth $0 after 10 years and negative after that — physically impossible. A $30,000 car losing 10% per year is exponential: V = 30,000(0.9)^t, worth about $10,460 after 10 years and always positive.
Items present a table and ask which model fits. Compute both differences and ratios: constant differences mean linear, constant ratios mean exponential.
Graph features
Every function of the form f(x) = ab^x with a > 0 has:
- y-intercept (0, a), since b⁰ = 1
- A horizontal asymptote at y = 0. The curve approaches the x-axis but never touches or crosses it, so there is no x-intercept.
- Range y > 0, and domain all real numbers
- Growth (b > 1): rises left to right, increasing ever more steeply, and approaches the asymptote to the far left
- Decay (0 < b < 1): falls left to right, dropping steeply then flattening, and approaches the asymptote to the far right
Selecting the right graph is exactly what the skill asks for. A graph that crosses the x-axis is not exponential. A graph that is a straight line is not exponential. A decay graph that flattens on the left rather than the right shows growth read backward.
Reading a rate from data
A bacterial culture is 400 cells at hour 0 and 1,350 at hour 3. Find the hourly growth factor. 400b³ = 1,350 → b³ = 3.375 → b = 3.375^(1/3) = 1.5 So the culture grows 50% per hour, and the model is P(t) = 400(1.5)^t.
The cube root step connects directly to the rational exponents of section 3.2: recovering a per-period factor from a multi-period change is a root extraction. Verify by forward computation: 400(1.5)³ = 400(3.375) = 1,350. ✓
A laptop worth $1,400 depreciates 12% per year. Which function models its value after t years, and what is its value after 5 years?
Which statement correctly distinguishes the graph of an exponential decay function from the graph of a decreasing linear function?
An investment of $8,000 earns 6% annual interest compounded monthly. Which expression gives its value after 4 years?