Exponential and Logarithmic Models Summary

Exponential and Logarithmic Models

• Four common types of mathematical models involving exponential and logarithmic functions:

  • Exponential Decay: $A = A_0e^{kt}$, where k < 0 is the decay rate.

  • Newton's Law of Cooling: $T(t) = Ae^{kt} + T<em>s$, where $T</em>s$ is the surrounding temperature.

  • Logistic Growth: $f(x) = \frac{c}{1 + Ae^{-Bx}}$.

Exponential Growth

• Model: $A(t) = A_0e^{kt}$, where:

  • $A(t)$ is the population at time t.

  • $A_0$ is the initial population size.

  • k is the growth rate.

  • t is time.

Example: Fruit Flies

• Initial setup to find $k$ includes two points:

  • After 2 days, 100 flies: $(2, 100)$

  • After 4 days, 300 flies: $(4, 300)$
    *Solving for k: $k = \frac{ln(3)}{2}$

  • Finding $A<em>0$: $A</em>0 = \frac{100}{3} = 33.33$

• Resulting function: $A(t) = 33.33e^{\frac{ln(3)}{2}t}$
  *Solving for $A(5)$ yields approximately 520 flies.

  • $A(5) = 33.33e^{\frac{ln(3)}{2}5} \approx 520$

Example: Bacteria doubling every two hours

• Equation: $200 = 100e^{k \cdot 2}$
  Solve for $k$: $k = \frac{ln(2)}{2}$. Function is $A(t) = A_0e^{\frac{ln(2)}{2}t}$.

Doubling Time

• Formula: $A(t) = A_0 \cdot 2^{\frac{t}{a}}$, where a is the doubling time.

Example: Bacterial culture doubling every 15 minutes

• Initial: $A_0 = 2000$, doubling time $a = 15$ minutes.

  • Amount of bacteria in 1 hour (60 minutes): $A(60) = 2000 \cdot 2^{\frac{60}{15}} = 32,000$

  • Amount after 6 hours (360 minutes): $A(360) = 2000 \cdot 2^{\frac{360}{15}} = 3.53 \times 10^{75}$

Exponential Decay

Example: Radioactive dye

• Initial dose: 10 mg. After 5 minutes, 6 mg remain. Detector sounds alarm if more than 2 mg are present.

  • Finding k: $6 = 10e^{k \cdot 5}$, so $k = \frac{ln(0.6)}{5}$.

  • Time until 2mg remains: $2 = 10e^{\frac{ln(0.6)}{5}t}$. Solving for t gives $t \approx 15.75$ minutes.

Half-Life

• Time for a quantity to reduce to half its original amount.
  Formula: $k = \frac{ln(0.5)}{t} = \frac{-ln(2)}{t}$

Example: Plutonium-239

• Half-life of 24,100 years. Amount remaining as a function of time:
  $A(t) = A<em>0e^{\frac{-ln(2)}{24100}t} = A</em>0 \cdot (\frac{1}{2})^{\frac{t}{24100}}$
  Alternate Half-Life Formula:

• $A(t) = A_0(\frac{1}{2})^{\frac{t}{a}}$, where a is the half-life.

Radiocarbon Dating

*Used to estimate the age of organic matter.

• Carbon-14 half-life: 5730 years.
  Formula for Carbon-14 remaining: $A = A_0e^{\frac{ln(0.5)}{5730}t}$
  Where,

• $A$ is the remaining amount of Carbon-14.

• $A_0$ is the intial amount of carbon-14 when the plant/animal began decaying.

Solving for age: $t = \frac{ln(r)}{-0.000121}$, where $r = \frac{A}{A_0}$. R is the percentage of carbon-14 in object to the percentage of carbon-14 in the air.

Example: Fossil with 35% carbon-14 compared to a living sample

Age: $t = \frac{ln(0.35)}{-0.000121} \approx 8,676$ years.

Newton's Law of Cooling

Formula: $T(t) = Ae^{kt} + T<em>s$, where: * T(t) is the temperature at time t. * $T</em>s$ is the surrounding air temperature.
* A is the difference between the initial temperature of the object and $T_s$.
* k is a constant.

Example: Coffee cooling

• Coffee starts at 180°F, room temperature is 76°F. After 5 minutes, coffee is 168°F.

  • $A = 180 - 76 = 104$. Initial function: $T(t) = 104e^{kt} + 76$.
    Solving for k using the information provided
    Substitute values into the equation

  • $168 = 104e^{k \cdot 5} + 76$ so $k = \frac{ln(\frac{92}{104})}{5}$.
    Function: $T(t) = 104e^{\frac{ln(\frac{92}{104})}{5}t} + 76$.
    Finding time for coffee to reach 155°F:
    Set $T(t)=155$.

  • $155 = 104e^{\frac{ln(\frac{92}{104})}{5}t} + 76$. Solving for t gives $t \approx 11.21$ minutes.

Logistic Growth

• Curve shape: Rapid growth that tapers off to an upper limit. Formula: $f(x) = \frac{c}{1 + Ae^{-Bx}}$, where:

  • $\frac{c}{1 + A}$ is the initial value.

  • c is the carrying capacity or limiting value.

  • B is a constant determined by the growth rate.

Example: Endangered species

• 100 animals released, carrying capacity of 1000, model: $P(t) = \frac{1000}{1 + 9e^{-0.1656t}}$

  • Population after 5 months: $P(5) = \frac{1000}{1 + 9e^{-0.1656 \cdot 5}} \approx 203$ animals.
    Number of months until the population is 500:
    Set $P(t)=500$ and Input

  • $500 = \frac{1000}{1 + 9e^{-0.1656t}}$. Solving for t gives $t \approx 13$ months.