Answer :
Let's detail the solution step-by-step:
### Part (a): Finding the Equation of the Exponential Function
The given data includes the year and the corresponding personal income. We aim to fit this data to an exponential function of the form:
[tex]\[ y = a \cdot e^{b \cdot x} \][/tex]
where:
- [tex]\( y \)[/tex] is the personal income,
- [tex]\( x \)[/tex] is the number of years after 1956,
- [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants to be determined.
After analyzing the provided data points and fitting them to this model, we find that the best-fitting parameters are:
[tex]\[ a = 769.257 \][/tex]
[tex]\[ b = 0.058 \][/tex]
Thus, the equation of the exponential function that models the data is:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot x} \][/tex]
### Part (b): Predicting Personal Income in 2006
We now use the established model to predict the personal income for the year 2006. Here, [tex]\( x \)[/tex] represents the number of years after 1956. Therefore, for the year 2006:
[tex]\[ x = 2006 - 1956 = 50 \][/tex]
Using the equation of the exponential function:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot 50} \][/tex]
The personal income in 2006 is calculated to be approximately:
[tex]\[ 13980.588 \, \text{billion dollars} \][/tex]
### Part (c): Predicting the Year When Personal Income Reaches [tex]$23$[/tex] Trillion
To determine the year when the personal income will reach \[tex]$23 trillion, we set \( y = 23000 \) (since 1 trillion = 1000 billion): \[ 23000 = 769.257 \cdot e^{0.058 \cdot x} \] We solve for \( x \): \[ \frac{23000}{769.257} = e^{0.058 \cdot x} \] \[ \ln\left(\frac{23000}{769.257}\right) = 0.058 \cdot x \] \[ x = \frac{\ln\left(\frac{23000}{769.257}\right)}{0.058} \] The value of \( x \) is approximately: \[ x \approx 58.583 \] Since \( x \) represents the number of years after 1956, the corresponding year is: \[ \text{Year} = 1956 + x \] \[ \text{Year} \approx 1956 + 58.583 \approx 2015 \] Therefore, according to the model, the total personal income will reach \$[/tex]23 trillion around the year 2015.
### Summary:
(a) The equation of the exponential function is:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot x} \][/tex]
(b) The total personal income in 2006 is predicted to be approximately:
[tex]\[ 13980.588 \, \text{billion dollars} \][/tex]
(c) The model predicts that the total personal income will reach \$23 trillion in the year:
[tex]\[ 2015 \][/tex]
### Part (a): Finding the Equation of the Exponential Function
The given data includes the year and the corresponding personal income. We aim to fit this data to an exponential function of the form:
[tex]\[ y = a \cdot e^{b \cdot x} \][/tex]
where:
- [tex]\( y \)[/tex] is the personal income,
- [tex]\( x \)[/tex] is the number of years after 1956,
- [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants to be determined.
After analyzing the provided data points and fitting them to this model, we find that the best-fitting parameters are:
[tex]\[ a = 769.257 \][/tex]
[tex]\[ b = 0.058 \][/tex]
Thus, the equation of the exponential function that models the data is:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot x} \][/tex]
### Part (b): Predicting Personal Income in 2006
We now use the established model to predict the personal income for the year 2006. Here, [tex]\( x \)[/tex] represents the number of years after 1956. Therefore, for the year 2006:
[tex]\[ x = 2006 - 1956 = 50 \][/tex]
Using the equation of the exponential function:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot 50} \][/tex]
The personal income in 2006 is calculated to be approximately:
[tex]\[ 13980.588 \, \text{billion dollars} \][/tex]
### Part (c): Predicting the Year When Personal Income Reaches [tex]$23$[/tex] Trillion
To determine the year when the personal income will reach \[tex]$23 trillion, we set \( y = 23000 \) (since 1 trillion = 1000 billion): \[ 23000 = 769.257 \cdot e^{0.058 \cdot x} \] We solve for \( x \): \[ \frac{23000}{769.257} = e^{0.058 \cdot x} \] \[ \ln\left(\frac{23000}{769.257}\right) = 0.058 \cdot x \] \[ x = \frac{\ln\left(\frac{23000}{769.257}\right)}{0.058} \] The value of \( x \) is approximately: \[ x \approx 58.583 \] Since \( x \) represents the number of years after 1956, the corresponding year is: \[ \text{Year} = 1956 + x \] \[ \text{Year} \approx 1956 + 58.583 \approx 2015 \] Therefore, according to the model, the total personal income will reach \$[/tex]23 trillion around the year 2015.
### Summary:
(a) The equation of the exponential function is:
[tex]\[ y = 769.257 \cdot e^{0.058 \cdot x} \][/tex]
(b) The total personal income in 2006 is predicted to be approximately:
[tex]\[ 13980.588 \, \text{billion dollars} \][/tex]
(c) The model predicts that the total personal income will reach \$23 trillion in the year:
[tex]\[ 2015 \][/tex]
