The general exponential growth model is:

$A(t) = A_0 e^{kt}$

where:

• $A(t)$ = amount at time $t$
• $A_0$ = initial amount (at $t = 0$)
• $k$ = growth rate constant ($k > 0$ for growth)
• $t$ = time
• $e \approx 2.71828$ (Euler's number)

Finding the Growth Rate $k$

If you know two data points, you can solve for $k$:

1. Substitute known values
2. Divide by $A_0$
3. Take natural log of both sides
4. Solve for $k$

Exponential Decay Model

Same formula, but $k < 0$ (negative):

$A(t) = A_0 e^{-kt}$

where $k > 0$ represents the decay rate.

Half-Life Formula

The half-life is the time it takes for half the substance to decay.

$A(t) = A_0\left(\frac{1}{2}\right)^{t/h}$

where $h$ is the half-life.

Relationship to $k$: $h = \frac{\ln(2)}{k}$

Compound Interest Models

Compound Interest (n times per year)

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

where:

• $P$ = principal (initial investment)
• $r$ = annual interest rate (as decimal)
• $n$ = number of times compounded per year
• $t$ = time in years

Continuous Compounding

$A = Pe^{rt}$

This is the limit as $n \to \infty$ (compounding infinitely often).

Doubling Time

The time it takes for a quantity to double:

$2A_0 = A_0e^{kt}$ $2 = e^{kt}$ $\ln(2) = kt$ $t = \frac{\ln(2)}{k}$

Doubling time: $t_d = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$

Population Growth Models

Unlimited Growth (Malthusian)

$P(t) = P_0e^{kt}$

Assumes unlimited resources (not realistic long-term).

Logistic Growth

$P(t) = \frac{L}{1 + Ce^{-kt}}$

where:

• $L$ = carrying capacity (maximum sustainable population)
• $C$ = constant based on initial conditions
• $k$ = growth rate

The population approaches $L$ as $t \to \infty$.

Newton's Law of Cooling

Temperature of an object over time:

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

where:

• $T(t)$ = temperature at time $t$
• $T_s$ = surrounding (ambient) temperature
• $T_0$ = initial temperature
• $k$ = cooling rate constant

Logarithmic Models

Some phenomena grow logarithmically:

$y = a + b\ln(x)$

Examples:

• Earthquake intensity (Richter scale)
• Sound intensity (decibels)
• pH scale (acidity)

Richter Scale

Earthquake magnitude: $M = \log\left(\frac{I}{I_0}\right)$

where $I$ is intensity and $I_0$ is reference intensity.

An increase of 1 on the Richter scale means 10 times more intense!

Solving Applied Problems

General Strategy:

1. Identify the type of model needed
2. Write the equation with known values
3. Solve for the unknown (often $k$ first, then answer the question)
4. Check if the answer makes sense in context

Step 1: Write the equation. $500 = 100e^{3k}$

Step 2: Divide by 100. $5 = e^{3k}$

Step 3: Take natural log. $\ln(5) = 3k$

Step 4: Solve for $k$. $k = \frac{\ln(5)}{3} \approx \frac{1.609}{3} \approx 0.536 \text{ per hour}$

Part 2: Find $P(5)$

$P(5) = 100e^{0.536(5)}$ $= 100e^{2.68}$ $\approx 100(14.58)$ $\approx 1458$

Answers:

• Growth rate: $k \approx 0.536$ per hour
• Population after 5 hours: approximately 1,458 bacteria

Step 1: Write the compound interest formula. $A = P\left(1 + \frac{r}{n}\right)^{nt}$

Step 2: Substitute values. $A = 5000\left(1 + \frac{0.06}{12}\right)^{12(10)}$

$= 5000\left(1 + 0.005\right)^{120}$

$= 5000(1.005)^{120}$

Step 3: Calculate. $(1.005)^{120} \approx 1.8194$

$A \approx 5000(1.8194) \approx 9097$

Answer: You will have approximately $9,097 after 10 years.

Step 1: Write the half-life formula. $A(t) = A_0\left(\frac{1}{2}\right)^{t/h}$

Step 2: Substitute known values. $0.25A_0 = A_0\left(\frac{1}{2}\right)^{t/5730}$

Step 3: Divide by $A_0$. $0.25 = \left(\frac{1}{2}\right)^{t/5730}$

Step 4: Recognize that $0.25 = \frac{1}{4} = \left(\frac{1}{2}\right)^2$. $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/5730}$

Step 5: Set exponents equal. $2 = \frac{t}{5730}$

Step 6: Solve for $t$. $t = 2(5730) = 11,460 \text{ years}$

Alternative method using logarithms:

$\ln(0.25) = \frac{t}{5730}\ln(0.5)$

$t = \frac{5730 \ln(0.25)}{\ln(0.5)} = 11,460 \text{ years}$

Answer: The fossil is approximately 11,460 years old.