Exponential Growth and Decay Applications

Vocabulary flashcards covering exponential growth, depreciation, and radioactive decay equations, evaluations, and logarithmic time-to-value solutions.

General Exponential Equation

The formula represented as $A = y_0(r)^t$, where $y_0$ is the initial amount, $r$ is the growth or decay rate factor, and $t$ is the time elapsed.

Bacteria Growth Equation (after 4 hours)

The expression $A = 500(2)^4$, modeling a population that begins at $500$ and doubles every hour evaluated after $4\,\text{hours}$.

Time to Reach 8,000 Bacteria

The duration $t = 4\,\text{hours}$, found by solving the equation $8000 = 500(2)^t$, which simplifies to $16 = (2)^t$ or $\frac{\log(16)}{\log(2)}$.

Car Depreciation Function

The function $A = 30000(0.85)^t$, which represents the value of a car purchased for \follower\$30000 that decreases in value by $15\%$ each year after $t\,\text{years}$.

Car Value After 5 Years

The depreciated car value calculated by evaluating $A = 30000(0.85)^5$, which yields \follower\$13311.16.

Time for Car Value to Fall Below $15,000

The time period of $4.27\,\text{years}$, determined by setting up $15000 = 30000(0.85)^t$ and solving $\frac{1}{2} = (0.85)^t$.

Radioactive Decay Equation

The exponential equation $A = 800(0.92)^t$, modeling a radioactive substance that starts with $800\,\text{grams}$ and loses $8\%$ of its mass each year.

Mass Remaining After 10 Years (Radioactive Substance)

The mass calculated by evaluating $A = 800(0.92)^{10}$, which equals $347.51\,\text{grams}$.

Time for Radioactive Mass to Drop Below 400 Grams

The time period of $8.31\,\text{years}$, obtained by solving $400 = 800(0.92)^t$, reducing to $\frac{1}{2} = (0.92)^t$ and computing $\frac{\log\left(\frac{1}{2}\right)}{\log(0.92)}$.