Problem 33 - Precalculus : Building Concepts and Connections

Question

The median price of a new home in the United States rose from \(\$ 123,000\) in 1990 to \(\$ 220,000\) in \(2004 .\) Find an exponential function \(P(t)=C e^{t t}\) that models the growth of housing prices, where \(t\) is the number of years since \(1990 .\) (Source: National Association of Home Builders)

Step-by-Step Solution

Answer

The exponential function modelling the growth of house prices over time since 1990 is:\[P(t) = 123,000 * e^{\frac{ln(\frac{220,000}{123,000})}{14} * t}\]

Step 1: Identifying Initial Conditions

Firstly, identify the initial conditions from the exercise. In 1990, the price of a house is $123,000. As 1990 denotes the start of our timeline, this is effectively the year 0 (t=0). As such, in the equation \(P(t) = C e^{kt}\), when \(t = 0\), \( P(t) = C\). So in this instance \(C = $123,000\) which represents the initial cost of a house.

Step 2: Setup Equation for 2004

Next, setup the exponential function to calculate the growth rate \(k\). In 2004, which is 14 years after 1990 (t=14), the price of a new house is $220,000. Substituting the known values (t, C, and P(t)) into the function gives: \[220,000 = 123,000 e^{14k}\]

Step 3: Solve for Growth Rate k

To calculate the exact growth rate \(k\), re-arrange the equation and solve for \(k\). Divide both sides by 123,000 and then take the natural log of both sides, rearranging once again to solve for \(k\):\[\frac{220,000}{123,000} = e^{14k}\]Taking natural logarithm,\[ln(\frac{220,000}{123,000}) = ln(e^{14k}) = 14k\]Solving for k,\[k = \frac{ln(\frac{220,000}{123,000})}{14}\]

Step 4: Formulate the Exponential Function

Finally, substitute values of \(C\) and \(k\) into the equation \(P(t) = C e^{kt}\), to formulate the exponential function that models the growth of house prices. That gives the equation for \(P(t)\) over years since 1990:

Key Concepts

Exponential Function

An exponential function is a mathematical expression used to describe processes that grow or decay at a constant rate. It often takes the form of \(P(t) = C e^{kt}\). The key components are:

• \(C\): The initial amount or starting value.
• \(e\): A constant approximately equal to 2.718, known as Euler's number.
• \(k\): The growth rate, determining how fast the function's output increases or decreases.
• \(t\): Time, often representing years, days, or any time unit.

This type of function is prevalent in fields like economics and biology, where quantities grow multiplicatively rather than additively.
It captures how things like populations, investments, and indeed, housing prices evolve over time. If \(k\) is positive, the function models growth; if negative, it models decay.

Housing Prices

Housing prices are pivotal in understanding economic health, often following trends over decades. These prices increase over time due to factors such as inflation, demand, and location desirability.
In the context of the exercise, we're examining the period from 1990 to 2004, where the prices reflected an upward trend.The price of a new home in 1990 was \\(123,000, rising to \\)220,000 by 2004.
Such a pattern of increase is well captured by an exponential growth model. It is a realistic representation due to:

• Compounded Growth: Price increases build upon previous increases, not remaining static.
• Sustained Demand: Continuous demand pushes prices upward over the years.

These prices don't just rise linearly but instead at a rate reflecting economic conditions and population pressures.

Initial Conditions

Initial conditions in a model set the starting point for the simulation or calculation. They are crucial because they impact the entire outcome of the model. For the housing price model, the initial conditions include:

• Year 1990 as Year 0: This is our baseline, with every subsequent year measured relative to 1990.
• Initial Home Price: The price of \$123,000 in 1990 is the starting value in our exponential function.

These initial conditions allow us to anchor the exponential function so that it accurately reflects real-world data.
By knowing where we begin (the price in 1990) and how much time has passed (14 years by 2004), we can employ our model to solve for the growth rate \(k\) and predict future prices accurately.
Without accurate initial conditions, predictions could become skewed, leading to ineffective forecasting.