Problem 19 - College Algebra

Question

Use this scenario: The population \(P\) of a koi pond over \(x\) months is modeled by the function \(P(x)=\frac{68}{1+16 e^{-0.28 x}}\). How many months will it take before there are 20 koi in the pond?

Step-by-Step Solution

Answer

It takes approximately 6.78 months.

Step 1: Set Up the Equation

We need to find when the population of the koi pond reaches 20. So we set the equation \( P(x) = 20 \). Thus, \( \frac{68}{1+16 e^{-0.28 x}} = 20 \).

Step 2: Solve for the Exponential Term

First, isolate the exponential term. Multiply both sides by \( 1 + 16 e^{-0.28 x} \) to get:\[68 = 20(1 + 16 e^{-0.28 x})\]Divide both sides by 20:\[\frac{68}{20} = 1 + 16 e^{-0.28 x}\]Simplify to:\[3.4 = 1 + 16 e^{-0.28 x}\]

Step 3: Isolate the Exponential

Subtract 1 from both sides to get:\[3.4 - 1 = 16 e^{-0.28 x}\]This simplifies to:\[2.4 = 16 e^{-0.28 x}\]

Step 4: Solve for the Exponential Value

Divide by 16 to isolate the exponential:\[\frac{2.4}{16} = e^{-0.28 x}\]This simplifies to:\[0.15 = e^{-0.28 x}\]

Step 5: Take the Natural Logarithm

Apply the natural logarithm \( \ln \) to both sides to solve for \( x \):\[\ln(0.15) = \ln(e^{-0.28 x})\]Using the property \( \ln(e^y) = y \), this simplifies to:\[\ln(0.15) = -0.28 x\]

Step 6: Solve for x

Finally, solve for \( x \) by dividing both sides by \(-0.28\):\[x = \frac{\ln(0.15)}{-0.28}\]Calculate \( x \) to get:\[x \approx \frac{-1.897}{-0.28} \approx 6.775\]So it takes approximately 6.78 months.

Key Concepts

Population Modeling

Population modeling involves using mathematical functions to represent the growth or decline of a population over time. In our scenario, the population of koi in a pond is modeled by the function \( P(x) = \frac{68}{1+16 e^{-0.28 x}} \). This type of equation is a logistic growth equation, which is commonly used for populations that grow quickly at first and then slow down as they approach a carrying capacity.

• **Initial Population and Growth**: At \( x = 0 \) months, the population will be the initial amount. We find this by evaluating \( P(0) \).
• **Carrying Capacity**: This is the maximum population size that the environment can sustain, which for the koi pond is 68 koi, the numerator of our function.

In this context, understanding population models helps in predicting how long it will take for the koi pond to reach a certain population. It is a crucial skill for managing animal populations and ecosystems.

Natural Logarithm

The natural logarithm is a mathematical function that is the inverse of the exponential function. It helps us solve equations where the variable is in the exponent, which is often the case in growth and decay problems.
In the scenario of finding the time it takes for the pond to reach 20 koi, once we have isolated the exponential term to \(0.15 = e^{-0.28 x} \), the next step involves taking the natural logarithm of both sides: \[\ln(0.15) = \ln(e^{-0.28 x}) \]The property \( \ln(e^y) = y \) simplifies the equation to :\( -0.28 x = \ln(0.15) \).
Natural logarithms are particularly useful because they allow us to "bring down" exponents, thereby making difficult equations more tractable. In the context of our problem, they help us solve for \( x \) directly, which represents time in months.

Solving Exponential Equations

Solving exponential equations involves finding the value of the variable that makes the equation true when the variable is part of an exponent. The exercise involves getting from a population function to the exact time it takes for a population of koi to reach a certain number.

• **Isolate the Exponential Part**: Move all other terms to one side to isolate the part of the equation that includes the exponent.
• **Apply the Natural Logarithm**: Use this step as it allows the exponent to be moved in front of the logarithm, solving for the time \( x \).
• **Rearrange and Solve**: Once the exponential term is isolated and you apply the natural logarithm, rearrange to solve for \( x \) by dividing both sides by the coefficient of \( x \).

In solving \( P(x) = 20 \), we ultimately found \( x \) by solving \( x = \frac{\ln(0.15)}{-0.28} \), which results in approximately 6.78 months. Mastering these steps is essential for anyone needing to model real-world scenarios involving exponential growth or decay.