Problem 65 - College Algebra

Question

The function \(f(x)=15,557+5259 \ln x\) models the average cost of a new car, \(f(x),\) in dollars, \(x\) years after 1989\. When was the average cost of a new car \(\$ 25,000 ?\)

Step-by-Step Solution

Answer

The average cost of a new car was $25000 about \(e^{9443/5259}\) years after 1989.

Step 1: Format the Equation

First, substitute the given average cost of $25,000 into the function: \(25,000 = 15,557 + 5259 \ln x\).

Step 2: Isolate the Logarithmic Term

To isolate the logarithmic term on one side of the equation, subtract 15,557 from both sides: \(25,000 - 15,557 = 5259 \ln x\). Simplifying gives, \(9,443 = 5259 \ln x\).

Step 3: Solve for the Logarithm

Next, divide both sides of the equation by 5259 to solve for \(\ln x\): \(\ln x = \frac{9,443}{5,259}\).

Step 4: Solve for \(x\)

Lastly, to solve for \(x\), use the definition of the natural logarithm, which states that the natural log of \(x\) is the power to which the base, \(e\), must be raised to produce \(x\). Therefore, \(x = e^{\frac{9,443}{5,259}}\).

Key Concepts

Logarithmic Functions

Logarithmic functions are a fundamental part of algebra. They are the inverses of exponential functions, meaning that they "undo" what exponentials do. In the context of our problem, the natural logarithm, denoted as \( \ln \), is used. \( \ln x \) refers to the power to which the base \( e \) (approximately 2.718) must be raised to get \( x \).

So, when the function involves \( 5259 \ln x \), this means we take \( \ln x \), multiply it by 5259, and then add it to the constant 15,557. This forms part of the model used to calculate the average car price.

• The base of the natural logarithm is \( e \).
• Logarithms are used to manage very large numbers due to their scale-reducing properties.
• Natural logarithms simplify calculations in growth-related equations.

In these equations, knowing how to manipulate logarithms is key to isolating variables and solving for unknown values. It is often a powerful tool in modeling real-world phenomena like costs and growth trends.

Solving Equations

Solving equations is a critical skill in college algebra. In the given exercise, you are essentially solving for \( x \) in the function \( f(x) = 15,557 + 5259 \ln x \). This involves a series of systematic steps that focus on isolating the variable of interest.

First, practical application begins by substituting \( 25,000 \) into the function as it equals the average cost of a new car. The aim is to determine the year corresponding to this cost.

• Begin by setting the function equal to the desired value.
• Isolate the logarithmic component by subtracting constants.
• Divide to remove coefficients and solve for the logarithmic term.

These steps allow you to work systematically from a more complex form of the equation into a simpler one that can be more straightforwardly interpreted and solved. Each stage narrows down the focus until you isolate the variable needed, in this case, \( x \). Such methodical solving techniques are pivotal in college algebra and beyond.

Mathematical Modeling

Mathematical modeling is the use of mathematical expressions to represent real-world situations. This skill is invaluable across many scientific and economic disciplines. The given equation \( f(x) = 15,557 + 5259 \ln x \) represents a mathematical model of car price growth over time, indicating its value as years after 1989 increase.

• Models can simplify complex systems into more understandable forms.
• They use available data to make predictions or to analyze trends.
• The effectiveness of a model depends on how accurately it captures relevant data.

In this specific model, the constant 15,557 represents the base price of a car in 1989, while \( 5259 \ln x \) represents how much additional value a car acquires over time, capturing trends in the cost as time progresses.
Mathematical models are not perfect, but by accurately using such models, predictions become more reliable and can be critical in planning and decision-making.

Exponential Functions

Exponential functions are mathematical expressions in which variables occur as exponents. They describe situations where a quantity grows or decays at a rate proportional to its current value, such as populations, investments, or in this case, costs.

While our main focus in the problem is the logarithmic function, solving for \( x \) from \( \ln x = y \) involves understanding that it implies \( x = e^y \), where \( e \) is the base of the natural logarithms.

• Exponential functions model growth that compounds over time.
• They are the inverses of logarithmic functions, providing a powerful symmetry in mathematics.
• Understanding exponentiation is vital for interpreting results from logarithmic equations.

This connection between logarithmic and exponential functions allows them to be used together effectively in many situations, from scientific research to financial forecasting, dramatically expanding their applicability in analysis and problem-solving.