Question
Total personal income of the country (in billions of dollars) for selected years from 1958 to 2003 is given in the table.
(a) These data can be modeled by an exponential function. Write the equation of this function, with x as the number of years after 1958.
(b) If this model is accurate, what will be the country's total personal income in 2008?
(c) In what year does the model predict the total personal income will reach $21 trillion?
Year
Personal Income
1958
414.2
1968
839.7
1978
2307.6
1988
4879.4
1998
8421.1
2003
10 comma 238.5
Question content area bottom
Part 1
(a) The equation of an exponential function that models the data is yequals
enter your response here.
Answers
GPT-4o mini
To model the total personal income data with an exponential function, we can use the general form of an exponential function:
\[
y = ab^x
\]
where:
- \(y\) is the total personal income,
- \(x\) is the number of years after 1958,
- \(a\) is the initial value (personal income in 1958), and
- \(b\) is the growth factor.
From the data provided:
- For the year 1958, the personal income is \(414.2\) billion dollars, which gives us \(a = 414.2\).
Now we will determine the growth factor \(b\). We can find \(b\) using the values from two known years. Using the years 1958 and 2003:
1. When \(x = 0\) (1958):
\[y = 414.2\]
2. When \(x = 45\) (2003):
\[y = 10238.5\]
We can set up the equation for the year 2003 as follows:
\[
10238.5 = 414.2 \cdot b^{45}
\]
To solve for \(b\):
\[
b^{45} = \frac{10238.5}{414.2}
\]
Calculating the right side:
\[
b^{45} \approx 24.7516
\]
To find \(b\), we will take the 45th root:
\[
b \approx (24.7516)^{\frac{1}{45}} \approx 1.0933
\]
Now, substituting back into the equation:
\[
y = 414.2 \cdot (1.0933)^x
\]
Thus, the exponential function modeling the data is:
\[
y = 414.2 \cdot (1.0933)^x
\]
---
### Part (b)
To find the country's total personal income in 2008, we first need to find \(x\) for 2008:
\[
x = 2008 - 1958 = 50
\]
Now we can substitute \(x = 50\) back into the equation:
\[
y = 414.2 \cdot (1.0933)^{50}
\]
Calculating \( (1.0933)^{50} \):
\[
(1.0933)^{50} \approx 13.288
\]
Now calculate \(y\):
\[
y \approx 414.2 \cdot 13.288 \approx 5501.7 \text{ billion dollars}
\]
So, if the model is accurate, the total personal income in 2008 is approximately **$5,501.7 billion**, or about **$5.5 trillion.**
---
### Part (c)
We need to determine the year when the total personal income \(y\) reaches $21 trillion, or 21,000 billion dollars. We will solve for \(x\):
\[
21000 = 414.2 \cdot (1.0933)^x
\]
Solving for \( (1.0933)^x \):
\[
(1.0933)^x = \frac{21000}{414.2} \approx 50.7083
\]
Now, take the logarithm:
\[
x \cdot \log(1.0933) = \log(50.7083)
\]
\[
x = \frac{\log(50.7083)}{\log(1.0933)} \approx \frac{1.7067}{0.0397} \approx 42.94
\]
Since \(x\) is approximately 42.94, round it to 43.
Now convert that back into the year:
\[
\text{Year} = 1958 + 43 = 2001
\]
Thus, the model predicts that the total personal income will reach **$21 trillion around the year 2001.**
\[
y = ab^x
\]
where:
- \(y\) is the total personal income,
- \(x\) is the number of years after 1958,
- \(a\) is the initial value (personal income in 1958), and
- \(b\) is the growth factor.
From the data provided:
- For the year 1958, the personal income is \(414.2\) billion dollars, which gives us \(a = 414.2\).
Now we will determine the growth factor \(b\). We can find \(b\) using the values from two known years. Using the years 1958 and 2003:
1. When \(x = 0\) (1958):
\[y = 414.2\]
2. When \(x = 45\) (2003):
\[y = 10238.5\]
We can set up the equation for the year 2003 as follows:
\[
10238.5 = 414.2 \cdot b^{45}
\]
To solve for \(b\):
\[
b^{45} = \frac{10238.5}{414.2}
\]
Calculating the right side:
\[
b^{45} \approx 24.7516
\]
To find \(b\), we will take the 45th root:
\[
b \approx (24.7516)^{\frac{1}{45}} \approx 1.0933
\]
Now, substituting back into the equation:
\[
y = 414.2 \cdot (1.0933)^x
\]
Thus, the exponential function modeling the data is:
\[
y = 414.2 \cdot (1.0933)^x
\]
---
### Part (b)
To find the country's total personal income in 2008, we first need to find \(x\) for 2008:
\[
x = 2008 - 1958 = 50
\]
Now we can substitute \(x = 50\) back into the equation:
\[
y = 414.2 \cdot (1.0933)^{50}
\]
Calculating \( (1.0933)^{50} \):
\[
(1.0933)^{50} \approx 13.288
\]
Now calculate \(y\):
\[
y \approx 414.2 \cdot 13.288 \approx 5501.7 \text{ billion dollars}
\]
So, if the model is accurate, the total personal income in 2008 is approximately **$5,501.7 billion**, or about **$5.5 trillion.**
---
### Part (c)
We need to determine the year when the total personal income \(y\) reaches $21 trillion, or 21,000 billion dollars. We will solve for \(x\):
\[
21000 = 414.2 \cdot (1.0933)^x
\]
Solving for \( (1.0933)^x \):
\[
(1.0933)^x = \frac{21000}{414.2} \approx 50.7083
\]
Now, take the logarithm:
\[
x \cdot \log(1.0933) = \log(50.7083)
\]
\[
x = \frac{\log(50.7083)}{\log(1.0933)} \approx \frac{1.7067}{0.0397} \approx 42.94
\]
Since \(x\) is approximately 42.94, round it to 43.
Now convert that back into the year:
\[
\text{Year} = 1958 + 43 = 2001
\]
Thus, the model predicts that the total personal income will reach **$21 trillion around the year 2001.**