Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in \(t\) years is modeled by the equation \(P(t)=\frac{1000}{1+9 e^{-0.6 t}}\). To the nearest tenth, how long will it take for the population to reach \(900 ?\)

Step-by-Step Solution

Answer

It takes approximately 7.3 years for the population to reach 900.

Step 1: Understand the Problem

We need to find how long it takes for the fish population, modeled by the equation \( P(t) = \frac{1000}{1+9 e^{-0.6 t}} \), to reach 900. This means we need to solve for \( t \) when \( P(t) = 900 \).

Step 2: Set Up the Equation

Substitute 900 for \( P(t) \) in the population equation:\[ 900 = \frac{1000}{1+9 e^{-0.6 t}} \]

Step 3: Solve the Equation

To solve \( 900 = \frac{1000}{1+9 e^{-0.6 t}} \), first multiply both sides by \( 1+9 e^{-0.6 t} \):\[ 900 (1+9 e^{-0.6 t}) = 1000 \]Next, distribute the 900:\[ 900 + 8100 e^{-0.6 t} = 1000 \]

Step 4: Isolate the Exponential Term

Subtract 900 from both sides to isolate the exponential term:\[ 8100 e^{-0.6 t} = 100 \]

Step 5: Solve for the Exponential

Divide both sides by 8100:\[ e^{-0.6 t} = \frac{100}{8100} \]Simplify the fraction:\[ e^{-0.6 t} = \frac{1}{81} \]

Step 6: Use Natural Logarithms

Take the natural logarithm (ln) of both sides:\[ \ln(e^{-0.6 t}) = \ln\left(\frac{1}{81}\right) \]Since \( \ln(e^x) = x \), this simplifies to:\[ -0.6 t = \ln\left(\frac{1}{81}\right) \]

Step 7: Solve for t

Divide both sides by \(-0.6\) to solve for \( t \):\[ t = \frac{\ln\left(\frac{1}{81}\right)}{-0.6} \]Calculate \( t \) using a calculator:\[ t \approx \frac{-4.394}{-0.6} \approx 7.3 \]

Step 8: Conclude the Solution

To the nearest tenth, it will take approximately 7.3 years for the population to reach 900.

Key Concepts

Population Modeling

Population modeling is a mathematical way to represent how a population, like that of a fish or animal species, grows over time. It uses specific formulas to predict changes based on various factors. In our fish farm scenario, the equation used is \[ P(t) = \frac{1000}{1+9e^{-0.6t}} \] Here, \( P(t) \) represents the population at time \( t \), 1000 is the carrying capacity (or maximum population), and the expression in the denominator captures the growth dynamics. The carrying capacity represents the upper limit of how many fish the environment can sustain.
The function also includes an exponential component, \( e^{-0.6t} \), which indicates the rate at which the population approaches the carrying capacity. This type of model is useful in predicting how quickly populations grow under ideal conditions. It helps farm managers and ecologists make informed decisions about resource allocation and environmental impacts.

Natural Logarithms

Natural logarithms, represented as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). They are particularly useful in dealing with exponential functions, especially in population growth models like the one in this exercise.
In step 6 of our solution, we used natural logarithms to solve an equation involving an exponential term. The original equation from the exponential step was \[ e^{-0.6t} = \frac{1}{81} \] Applying natural logarithms to both sides transformed this into:
\[ \ln(e^{-0.6t}) = \ln\left(\frac{1}{81}\right) \] Using the property \( \ln(e^x) = x \), we simplified this equation to:
\[ -0.6t = \ln\left(\frac{1}{81}\right) \] This simplification allows us to solve for \( t \), demonstrating the power of logarithms in handling equations where time or rates of growth must be isolated. Understanding how to apply natural logarithms is essential for working with exponential models effectively.

Graphing Calculator Usage

Graphing calculators are invaluable tools for visualizing and solving complex mathematical equations, like our population model. In computations like the one in this exercise, they help verify answers and provide a visual understanding of how variables interact.
A graphing calculator can graph the function \[ P(t) = \frac{1000}{1+9e^{-0.6t}} \] and show how the population approaches the carrying capacity over time. To do this, input the equation into the calculator and view the graph to see the growth curve of the population. This visual can make it easier to understand concepts of growth rate and equilibrium states.
Additionally, a graphing calculator can handle calculations involving natural logarithms directly, allowing you to enter equations like \[ t = \frac{\ln\left(\frac{1}{81}\right)}{-0.6} \] Efficient use of these tools can save time and increase accuracy in your calculations, allowing learners to focus on understanding the concepts rather than getting bogged down by arithmetic.