Exponential Growth and Decay

Exponential growth and decay apply to substances or populations where the rate of change at any time $t$ is proportional to the amount present.
The general mathematical model is given by the equation: $y = Ce^{kt}$
$C$ represents the initial value (when $t = 0$ ).
$k$ is the constant of proportionality.
Growth is represented by k > 0.
Decay is represented by k < 0.

Establish two sets of conditions using the provided information for variables $y$ and $t$ .
Substitute these conditions into the model $y = Ce^{kt}$ to solve for the constants $C$ and $k$ . If one condition involves $t = 0$ , use it first to find $C$ .
Use the completed model to calculate the final answer.

Radioactive decay is quantified using half-life, which is the time required for half of the atoms in a sample of radioactive material to decay.

Scenario: A population follows an exponential growth model.
Conditions:
- After 2 days ( $t = 2$ ): 100 flies.
- After 4 days ( $t = 4$ ): 300 flies.
Model Solution:
- Solving for the constants yields $C \approx 33$ and $k \approx 0.5493$ .
- The resulting model is $y = 33e^{0.5493t}$ .
Prediction: After 5 days ( $t = 5$ ), the population is estimated at: $y = 33e^{0.5493(5)} \approx 514\, \text{flies}$

While $e$ is the standard base, exponential growth can be modeled with any base $a$ : $y = Ca^{bt}$
This relationship can be expressed using base $e$ as: $y = Ce^{(\ln(a))bt}$

Scenario: A manufacturer notices a drop in sales after national advertising is discontinued.
Conditions:
- Initial sales ( $t = 0$ ): 100,000 MP3 players.
- After 4 months ( $t = 4$ ): 80,000 MP3 players.
Model Solution:
- Given $t = 0$ , $C = 100,000$ .
- Substitution of the second condition identifies the decay constant $k$ .
Prediction: After four additional months ($t = 8$), the expected sales drop to: $y = 64,000\, \text{MP3 players}$