Total personal income of the country​ (in billions of​ dollars) for selected years from 1958 to 2003 is given in the table.

Question

​(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
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Part 1
​(a) The equation of an exponential function that models the data is yequals
  
enter your response here.

Answer 1

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:

\[ y = 414.2 \cdot (1.0933)^x \]

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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.

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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.