Exponential Growth and Decay Models

Exponential functions are perfect for modeling real-world situations where a quantity increases or decreases by the same percentage over time. This includes population growth, compound interest, depreciation of a car's value, and radioactive decay.

The Standard Formula

Both growth and decay can be modeled using the same basic exponential function:

$y = a \cdot b^x$

Here is what each part of the formula represents:

• $y$: The final amount after time $x$.
• $a$: The initial amount (the starting value when $x = 0$).
• $b$: The growth or decay factor (the multiplier).
• $x$: The amount of time that has passed (or the number of intervals).

Identifying Growth vs. Decay

You can easily tell if an equation represents growth or decay by looking at the base, $b$.

• Exponential Growth ($b > 1$): The quantity is increasing. The growth factor is found by adding the percentage rate (as a decimal) to 1: $b = 1 + r$.
• Exponential Decay ($0 < b < 1$): The quantity is decreasing. The decay factor is found by subtracting the percentage rate (as a decimal) from 1: $b = 1 - r$.

Example Problems

Example 1: Identifying the Model

Determine whether $y = 2(0.85)^x$ represents growth or decay.

Look at the base of the exponent, which is $0.85$. Since $0 < 0.85 < 1$, the base is less than one. Therefore, this equation represents exponential decay. (Specifically, it represents a 15% decrease per time period, because $1 - 0.15 = 0.85$).

Example 2: Modeling Population Growth

A population of 500 grows at 3% per year. Write the model and find the population after 10 years.

1. Identify the variables:
   • Initial amount, $a = 500$
   • Growth rate, $r = 3\% = 0.03$
   • Growth factor, $b = 1 + 0.03 = 1.03$
2. Write the model: Substitute $a$ and $b$ into the standard formula: $y = 500(1.03)^x$
3. Solve for $x = 10$: Plug in 10 for the number of years: $y = 500(1.03)^{10}$ $y \approx 500(1.3439)$ $y \approx 671.96$

After 10 years, the population will be approximately 672.