Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, use this scenario: The population \(P\) of a koi pond over \(x\) months is modeled by the function \(P(x)=\frac{68}{1+16 e^{-0.28 x}}\) Use the intersect feature to approximate the number of months it will take before the population of the pond reaches half its carrying capacity.

Step-by-Step Solution

Answer

It takes approximately 9.9 months.

Step 1: Determine Carrying Capacity

The function provided is a logistic growth model. The carrying capacity is the maximum population the environment can sustain. In the equation \( P(x) = \frac{68}{1 + 16e^{-0.28x}} \), the carrying capacity \( K \) is the numerator of the fraction, which is \( K = 68 \).

Step 2: Calculate Half of Carrying Capacity

Half of the carrying capacity is \( \frac{68}{2} = 34 \). We need to find the number of months \( x \) that will make the population reach 34.

Step 3: Set up the Equation to Solve for x

We want \( P(x) = 34 \). Therefore, set \( \frac{68}{1 + 16e^{-0.28x}} = 34 \).

Step 4: Solve the Equation Algebraically

First, multiply both sides of the equation by \( 1 + 16e^{-0.28x} \) to eliminate the denominator:\[ 68 = 34(1 + 16e^{-0.28x}) \] Expand the right side:\[ 68 = 34 + 544e^{-0.28x} \]Subtract 34 from both sides:\[ 34 = 544e^{-0.28x} \]

Step 5: Isolate the Exponential Term

Divide both sides by 544:\[ \frac{34}{544} = e^{-0.28x} \]Simplify the fraction:\[ \frac{34}{544} = 0.0625 \]So, \( e^{-0.28x} = 0.0625 \).

Step 6: Solve for x Using Natural Logarithms

Take the natural logarithm of both sides:\[ \ln(e^{-0.28x}) = \ln(0.0625) \]This simplifies to:\[ -0.28x = \ln(0.0625) \]Thus, \( x = \frac{\ln(0.0625)}{-0.28} \).

Step 7: Calculate x

Compute \( x = \frac{\ln(0.0625)}{-0.28} \). Use a calculator to find \( \ln(0.0625) \approx -2.7726 \). Therefore, \[ x \approx \frac{-2.7726}{-0.28} \approx 9.902 \] months.

Key Concepts

Understanding Carrying Capacity in Logistic Growth Models

The concept of carrying capacity is central to understanding logistic growth models. In biology and ecology, carrying capacity refers to the maximum population size of a species that an environment can sustain indefinitely. In other words, it is the limit beyond which a population cannot grow due to constraints such as limited resources, space, and other environmental factors.
In the context of the logistic growth model equation given in the problem, the carrying capacity is represented by the numerator in the formula: \[ P(x) = \frac{K}{1 + ae^{-bx}} \]where \( K \) is the carrying capacity. This value defines the upper limit of the population.

• In the equation given, \( K \) is clearly specified as 68. So, the pond can support up to 68 koi fish at maximum.
• As a population nears this limit, its growth rate slows, compared to the rapid growth seen when the population is much lower.
• Reaching half the carrying capacity, for example, is a common point of interest in exercises since it illustrates significant mid-point growth during change.

Understanding carrying capacity helps us not only in solving the mathematical problem but also in assessing the ecological balance within an ecosystem.

The Role of Natural Logarithms in Solving Exponential Equations

Natural logarithms, denoted as \( \ln \), are often used in solving problems with exponential growth or decay. In our problem, we encounter the exponential equation:\[ e^{-0.28x} = 0.0625 \]To solve for \( x \), which represents time in months in this scenario, natural logarithms play a pivotal role.

• Natural logarithms convert multiplicative operations into additive ones, making equations easier to handle mathematically.
• The natural logarithm of the exponential function has a particularly simple form: \( \ln(e^{y}) = y \), which simplifies calculations significantly.
• By taking the natural logarithm of both sides, we can solve equations like \( -0.28x = \ln(0.0625) \), greatly simplifying the process of solving for unknowns in exponential equations.

This transformation is crucial for breaking down complex exponential terms into manageable linear equations, facilitating the computation of time-based variables in growth models.

Working with Exponential Equations in Population Models

Exponential equations are fundamental to models that describe growth or decay, and they appear frequently within natural phenomena such as population dynamics, radioactive decay, and interest calculations.
In logistic growth models like the one in our exercise, the equation:\[ P(x) = \frac{68}{1 + 16e^{-0.28x}} \]has an exponential component \( e^{-0.28x} \), which describes how quickly the population grows initially.
Exponential equations often involve terms that can cause rapid changes in response to small differences in the exponent.

• The base of the natural logarithm, \( e \), is approximately 2.718, and it is central to calculating changes related to exponential growth.
• In our model, the exponent \(-0.28x\) impacts the speed of the population's change—in this case, how fast the koi population grows before reaching a slow and steady state as it approaches carrying capacity.
• Solving exponential equations usually involves isolating the exponential term and using logarithms to solve for the unknown variable, as detailed in the steps of the solution.

Mastering exponential equations provides a deeper understanding of growth patterns and an ability to predict future changes in populations, essential skills in various scientific and financial fields.