Lesson 6.7 : Exponential and Logarithmic Models

New Concept

Learn to model real-world phenomena using exponential and logarithmic functions. We'll explore concepts like growth (doubling time), decay (half-life), Newton's Law of Cooling, and growth with limits using logistic models.

What’s next

Next, you'll tackle interactive examples and practice cards on growth, decay, and logistic models to build your skills.

Property

In the case of rapid growth, we may choose the exponential growth function: $y = A_0 e^{kt}$, where $A_0$ is equal to the value at time zero, $e$ is Euler’s constant, and $k$ is a positive constant that determines the rate (percentage) of growth. For exponential decay, $k$ is a negative constant that determines the rate of decay.

An exponential function with the form $y = A_0 e^{kt}$ has the following characteristics:

• one-to-one function
• horizontal asymptote: $y = 0$
• domain: $(-\infty, \infty)$
• range: $(0, \infty)$
• x intercept: none
• y-intercept: $(0, A_0)$
• increasing if $k > 0$
• decreasing if $k < 0$

Examples

• A colony of ants starts with 200 ants and grows according to the function $A(t) = 200e^{0.05t}$. Since $k=0.05 > 0$, this represents exponential growth. The initial population is $A_0 = 200$.

Property

Half-life is the length of time it takes an exponentially decaying quantity to decrease to half its original amount. The formula for half-life $t$ is derived from $\frac{1}{2}A_0 = A_0 e^{kt}$ and is given by:

Examples

• The half-life of Bismuth-210 is 5 days. To find its decay rate $k$, we use the formula $k = \frac{-\ln(2)}{t} = \frac{-\ln(2)}{5} \approx -0.1386$. The decay model is $A(t) = A_0e^{-0.1386t}$.

• A wooden spear found at an archeological site contains 40% of its original carbon-14. Its age is calculated as $t = \frac{\ln(0.40)}{-0.000121} \approx 7575$ years old.

Property

For growing quantities, we might want to find out how long it takes for a quantity to double. The time it takes for a quantity to double is called the doubling time.

Given the basic exponential growth equation $A = A_0 e^{kt}$, doubling time can be found by solving for when the original quantity has doubled, that is, by solving $2A_0 = A_0 e^{kt}$. The formula is:

Examples

• A city's population is growing at a continuous rate of 3% per year ($k=0.03$). The time it will take for the population to double is $t = \frac{\ln 2}{0.03} \approx 23.1$ years.

Property

The temperature of an object, $T$, in surrounding air with temperature $T_s$ will behave according to the formula

where

• $t$ is time
• $A$ is the difference between the initial temperature of the object and the surroundings
• $k$ is a constant, the continuous rate of cooling of the object

To apply the law, set $T_s$ to the ambient temperature, then use given values to find the parameters $A$ and $k$.

Examples

• A baked pie at $200^\circ$F is placed in a $70^\circ$F room. Here, $T_s = 70$. The initial temperature difference is $A = 200 - 70 = 130$. The cooling model is $T(t) = 130e^{kt} + 70$.

• A bottle of water at $40^\circ$F is moved to a room at $75^\circ$F. After 20 minutes, it warms to $50^\circ$F. We have $T_s = 75$ and $A = 40 - 75 = -35$. We solve $50 = -35e^{k \cdot 20} + 75$ to find $k$.

Property

The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. The logistic growth model is

where

• $\frac{c}{1+a}$ is the initial value
• $c$ is the carrying capacity, or limiting value
• $b$ is a constant determined by the rate of growth.

Examples

• A new social media app's user base is modeled by $f(t) = \frac{2,000,000}{1 + 500e^{-0.8t}}$. The carrying capacity is $c = 2,000,000$ users. The initial number of users was $\frac{2,000,000}{1+500} \approx 3992$.

• A deer population in a forest is limited to 500. It starts with 20 deer and the growth constant is $b=0.4$. Here, $c=500$. We find $a$ from $20 = \frac{500}{1+a}$, so $a=24$. The model is $P(t) = \frac{500}{1 + 24e^{-0.4t}}$.

Property

While powers and logarithms of any base can be used in modeling, the two most common bases are 10 and $e$. We can use laws of exponents and laws of logarithms to change any base to base $e$.

Given a model with the form $y = ab^x$, change it to the form $y = A_0 e^{kx}$.

1. Rewrite $y = ab^x$ as $y = a(e^{\ln b})^x$.
2. Use the power rule of exponents to rewrite $y$ as $y = ae^{x \ln(b)}$.
3. Note that $a = A_0$ and $k = \ln(b)$ in the equation $y = A_0 e^{kx}$.

Examples

• To change $y = 4(1.2)^x$ to base $e$, we identify $a=4$ and $b=1.2$. The new form is $y = 4e^{(\ln 1.2)x}$. Since $\ln(1.2) \approx 0.1823$, the function is $y = 4e^{0.1823x}$.