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Sagot :
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.
[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.
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