Elias purchases a home for [tex]$\$[/tex]38,900[tex]$. The value of the home, in thousands of dollars, since his purchase is shown in the table.

\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Years since \\
Purchase
\end{tabular} & \begin{tabular}{c}
Value \\
(thousands of $[/tex]\[tex]$)
\end{tabular} \\
\hline
0 & 38.9 \\
\hline
5 & 62.4 \\
\hline
10 & 89.3 \\
\hline
15 & 145.2 \\
\hline
20 & 210.8 \\
\hline
25 & 326.5 \\
\hline
\end{tabular}

Find an exponential function that models the data. Round numerical values to the nearest hundredth. Let $x$ be the number of years since the purchase.

The function $f(x) = \square ( )^x$ models the data.

Use the model to predict the home's value.

After 12 years, the home's value will be about $\$[/tex] (thousands).

After 35 years, the home's value will be about [tex]$\$[/tex]$ (thousands).



Answer :

To solve the problem of modeling the value of Elias's home, we need to find an exponential function of the form:

[tex]\[ f(x) = a \cdot b^x \][/tex]

where:
- [tex]\( x \)[/tex] is the number of years since Elias purchased the home,
- [tex]\( a \)[/tex] is the initial value coefficient,
- [tex]\( b \)[/tex] is the base of the exponential function representing the growth rate over time.

Given the data from the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Years since purchase} & \text{Value (thousands of dollars)} \\ \hline 0 & 38.9 \\ \hline 5 & 62.4 \\ \hline 10 & 89.3 \\ \hline 15 & 145.2 \\ \hline 20 & 210.8 \\ \hline 25 & 326.5 \\ \hline \end{array} \][/tex]

To find the exponential function that fits this data, we process the data through the following steps:

1. Apply a logarithmic transformation to the values to linearize the exponential relationship.
2. Use linear regression to find the slope and intercept of the line fit to the log-transformed values.
3. Transform the slope and intercept back to the corresponding parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex] of the exponential function.

From the data processing, we obtain:

1. The initial value coefficient [tex]\(\mathbf{a = 39.59}\)[/tex]. This represents the value of the home at the initial time (0 years).
2. The base of the exponential function [tex]\(\mathbf{b = 1.09}\)[/tex]. This represents the growth rate per year.

Thus, the exponential function that models the data is:

[tex]\[ f(x) = 39.59 \cdot (1.09)^x \][/tex]

To predict the value of the home after 12 years, we substitute [tex]\( x = 12 \)[/tex] into the exponential function:

[tex]\[ f(12) = 39.59 \cdot (1.09)^{12} \][/tex]
[tex]\[ f(12) \approx 109.05 \][/tex]

After 12 years, the home's value will be about [tex]\(\$109.05\)[/tex] thousand.

To predict the value of the home after 35 years, we substitute [tex]\( x = 35 \)[/tex] into the exponential function:

[tex]\[ f(35) = 39.59 \cdot (1.09)^{35} \][/tex]
[tex]\[ f(35) \approx 760.28 \][/tex]

After 35 years, the home's value will be about [tex]\(\$760.28\)[/tex] thousand.

Therefore, the final answers are:

1. The exponential function that models the data is:

[tex]\[ f(x) = 39.59 \cdot (1.09)^x \][/tex]

2. After 12 years, the home's value will be about [tex]\(\$109.05\)[/tex] thousand.

3. After 35 years, the home's value will be about [tex]\(\$760.28\)[/tex] thousand.