Problem 14 - College Algebra with Corequisite Support

Question

A logistic model is given by the equation \(P(t)=\frac{90}{1+5 e^{-0.42 t}} .\) To the nearest hundredth, for what value of \(t\) does \(P(t)=45 ?\)

Step-by-Step Solution

Answer

For \( t \approx 3.91 \), \( P(t) = 45 \).

Step 1: Set up the Equation

We are given the logistic model \( P(t) = \frac{90}{1+5e^{-0.42t}} \) and need to find the value of \( t \) when \( P(t) = 45 \). Let's set \( P(t) = 45 \) in the equation: \( 45 = \frac{90}{1+5e^{-0.42t}} \).

Step 2: Solve for the Denominator

To eliminate the fraction, multiply both sides by the denominator: \( 45(1+5e^{-0.42t}) = 90 \). Simplify it to \( 45 + 225e^{-0.42t} = 90 \).

Step 3: Isolate the Exponential Term

Subtract 45 from both sides to isolate the exponential term: \( 225e^{-0.42t} = 45 \).

Step 4: Divide to Simplify

Divide both sides by 225 to solve for \( e^{-0.42t} \): \( e^{-0.42t} = \frac{45}{225} \). Simplify the fraction to \( e^{-0.42t} = 0.2 \).

Step 5: Take the Natural Logarithm

Apply the natural logarithm (ln) to both sides to eliminate the exponential function: \( \ln(e^{-0.42t}) = \ln(0.2) \). This simplifies to \( -0.42t = \ln(0.2) \).

Step 6: Solve for t

Divide both sides by -0.42 to solve for \( t \): \( t = \frac{\ln(0.2)}{-0.42} \). Calculate this value using a calculator: \( t \approx 3.91 \).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is an essential mathematical function when working with exponential equations, such as those found in logistic growth models. It is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. When you have an equation involving exponentials, taking the natural logarithm can help you solve for unknown variables.

In our logistic growth model, the equation included an exponential function, \( e^{-0.42t} \). To isolate \( t \), we used the natural logarithm to remove the \( e \). Applying \( \ln \) to both sides allows us to simplify \( \ln(e^{-0.42t}) \) into \( -0.42t \). This is because the logarithm of a power means multiplying the logarithm by the exponent.

• Natural logarithms are commonly used in continuous growth models, like populations or radioactive decay.
• They are crucial for transforming exponential equations into a solvable form.
• The expression \( \ln(e^{x}) = x \) is a fundamental property that helps in simplifying equations.

Exponential Equations

Exponential equations involve variables in the exponent and appear frequently in scientific models, including the logistic growth model in our exercise. These equations describe processes that increase or decrease at rates proportional to their current size, such as population growth, interest in finance, or chemical reactions.

The key component of an exponential equation is the base of the exponent, usually \( e \), to represent continuous growth or decay. In the logistic model \( P(t) = \frac{90}{1+5e^{-0.42t}} \), the exponential equation is embedded in the denominator.

To solve this kind of equation, isolate the exponential term. Often, you will:

• Multiply and divide to move constants around.
• Use logarithms to take the exponent down to a linear term.
• Simplify the expressions carefully step by step.

In the solution process, we needed to convert \( e^{-0.42t} = 0.2 \) so that \( t \) could be explicitly solved. Understanding these maneuvers is vital in demystifying exponential behaviors in complex models.

Solving Equations

Solving equations is a core skill in mathematics, enabling us to find unknown values that satisfy a given mathematical statement. This skill is especially significant in the realm of algebra, where we deal with varying complexity from simple linear equations to more involved exponential and logarithmic equations.

For our logistic model, solving the equation involves:

• Setting equations by substituting known values, \( P(t) = 45 \).
• Manipulating the equation using algebraic rules to isolate the exponential component.
• Applying logarithms to seamlessly handle exponential terms, turning them into linear expressions.
• Rearranging and simplifying to find \( t \), our desired variable.

Breaking down each step, from setting the equation up to arriving at \( t \approx 3.91 \), not only solves the problem but also strengthens problem-solving strategies. Understanding the methodical approach underpinning this process aids students in tackling a variety of mathematical scenarios beyond logistics.