Problem 82 - Precalculus

Question

If the number of new model Honda Accord hybrids purchased in North America is given by \(N=\frac{100,000}{1+10 e^{-2 t}},\) where \(t\) is the number of weeks after Honda releases the new model, how many weeks will it take after the release until there are 50,000 Honda hybrids from that batch on the road?

Step-by-Step Solution

Answer

It will take approximately 2 weeks for 50,000 hybrids to be on the road.

Step 1: Understand the Problem

The problem involves finding the time, denoted as \( t \), in weeks when the number \( N \) of Honda Accord hybrids reaches 50,000. The function given is \( N = \frac{100,000}{1+10 e^{-2 t}} \). We need to solve this equation for \( t \) when \( N = 50,000 \).

Step 2: Set the Equation

Set the equation \( \frac{100,000}{1+10 e^{-2 t}} = 50,000 \). This sets up the requirement that 50,000 cars are on the road.

Step 3: Simplify the Equation

Multiply both sides of the equation by \( 1 + 10e^{-2t} \) to eliminate the fraction: \[ 100,000 = 50,000(1 + 10e^{-2t}) \].

Step 4: Distribute and Rearrange

Distribute the 50,000 on the right-hand side to get: \[ 100,000 = 50,000 + 500,000e^{-2t} \].Subtract 50,000 from both sides to obtain: \[ 50,000 = 500,000e^{-2t} \].

Step 5: Solve for the Exponential Term

Divide both sides by 500,000 to solve for the exponential term:\[ \frac{50,000}{500,000} = e^{-2t} \].This simplifies to: \[ e^{-2t} = 0.1 \].

Step 6: Apply Natural Logarithm

Take the natural logarithm of both sides to solve for \( t \): \[ \ln(e^{-2t}) = \ln(0.1) \]. Simplifying gives: \[ -2t = \ln(0.1) \].

Step 7: Solve for t

Solve for \( t \) by dividing both sides by \(-2\): \[ t = -\frac{\ln(0.1)}{2} \]. Calculate the value: \( \ln(0.1) \approx -2.3026 \), and then divide by \(-2\) to find \( t \approx 1.1513 \).

Step 8: Finalize the Answer

Round \( t \) to the nearest week, as \( t \approx 1.1513 \). Therefore, it will take approximately 2 weeks.

Key Concepts

Natural Logarithms

Natural logarithms are a fundamental part of mathematics used particularly in cases involving exponential functions. The natural logarithm, written as \( \ln(x) \), is the inverse function of an exponential \( e^x \). This means when you have \( e^x = y \), you can find \( x \) by calculating \( \ln(y) \).

• Natural logs are base \( e \), where \( e \approx 2.71828 \), a mathematical constant.
• This operation helps when working with exponential decay or growth scenarios.

In the given problem, the function describing the number of cars \( N \) is dependent on an exponential decay term \( e^{-2t} \). To solve for \( t \), the natural logarithm is employed. By taking the natural log of both sides of the equation \( e^{-2t} = 0.1 \), we effectively "bring down" the exponent, making it easier to solve for \( t \). This simplifies the expression to \( -2t = \ln(0.1) \), allowing straightforward solving for \( t \).
Natural logarithms simplify complex equations, making it possible to solve them using arithmetic. They are very useful in contexts where variables are exponents.

Solving Equations

Solving equations, especially those involving exponential functions, often requires breaking them down into simpler steps. In the exercise, we start with the equation \( \frac{100,000}{1+10 e^{-2 t}} = 50,000 \) and manipulate it to isolate \( t \). Here's the step-by-step approach:

• First, eliminate fractions by multiplying both sides by \( 1 + 10e^{-2t} \).
• Next, distribute and simplify the equation to extract the exponential term.
• After isolating \( e^{-2t} \), logarithmic operations can be applied to both sides.

This method demonstrates how we isolate variables, especially those within exponents, to simplify and solve for desired values. Each step, such as multiplying through to eliminate denominators, is crucial as it leads toward the ultimate goal of simplifying the complexity of the equation. In general, solving equations involves a strategic process where simplifying and rearranging terms leads to an understandable solution.

Function Analysis

Function analysis is an essential part of understanding how changes in variables affect the whole equation. The original function \( N = \frac{100,000}{1+10 e^{-2 t}} \) describes how many Honda Accords are on the road depending on weeks passed. By analyzing this function, we can predict future behavior.

• The function shows a type of growth limited by the maximum value of 100,000.
• This is indicative of a saturation point or asymptote, typical in logistic growth models.

As time \( t \) increases, the growth of \( N \) slows, reflecting real-world models where resources or populations can't grow indefinitely. Understanding such features allows us to grasp the relationships between time, growth, and limits without complex calculations.
Function analysis also involves checking the behavior at extreme values. As \( t \rightarrow \infty \), \( e^{-2t} \rightarrow 0 \), leading \( N \) to approach its maximum value of 100,000. These insights help in anticipating the long-term expectations of the model, offering a comprehensive understanding of the situation.