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.**