### Predicting Year China’s Automobile and Parts Exports Reach $11.1 Billion#### Problem Statement:The value of China's exports of automobiles and parts (in billions of dollars) is approximately given by the function:\[ f(x) = 1.8208e^{0.3387x} \]where $ x = 0 $ corresponds to the year 1998.#### Question:In what year did/will the exports reach $11.1 billion?**Instructions:** Give your answer as the year, with at least one decimal place.### Solution Approach:To find the year when the exports reach $11.1 billion, set $ f(x) $ equal to 11.1 and solve for $ x $:\[11.1 = 1.8208e^{0.3387x}\]**Steps to Solve:**1. Divide both sides by 1.8208: \[ \frac{11.1}{1.8208} = e^{0.3387x} \]2. Calculate the division: \[ 6.096 \approx e^{0.3387x} \]3. Take the natural logarithm (ln) of both sides to solve for $ x $: \[ \ln(6.096) = 0.3387x \]4. Simplify using the properties of logarithms: \[ x = \frac{\ln(6.096)}{0.3387} \]5. Compute the value of $ x $.6. Add the computed value of $ x $ to the base year (1998) to find the approximate year.You can use a calculator or a computational tool to find the exact value.### Graphical Explanation:If there were a graph depicting this function, it would show an exponential growth curve starting from the year 1998. The x-axis would represent the years, starting from 1998, and the y-axis would represent the export value in billions of dollars. The point where the curve intersects the horizontal line at $11.1 billion would indicate the year you need to determine.### Additional Resources:For further guidance, you can refer to the video by clicking on the “Question Help” link provided in the original material or use other educational resources available on exponential growth and logarithmic functions.### Answer

The value of China's export of automobiles and parts is approximated by the , where corresponds to…

The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x) = 1.8208e-3387 where x = 0 corresponds to 1998. In what year did/will the exports reach $11.1 billion? 3

The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x) = 1.8208e-3387 where x = 0 corresponds to 1998. In what year did/will the exports reach $11.1 billion? 3

Algebra and Trigonometry (6th Edition)

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Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

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Question

### Predicting Year China’s Automobile and Parts Exports Reach $11.1 Billion

#### Problem Statement:
The value of China's exports of automobiles and parts (in billions of dollars) is approximately given by the function:

\[ f(x) = 1.8208e^{0.3387x} \]

where $ x = 0 $ corresponds to the year 1998.

#### Question:
In what year did/will the exports reach $11.1 billion?

**Instructions:**
Give your answer as the year, with at least one decimal place.

### Solution Approach:

To find the year when the exports reach $11.1 billion, set $ f(x) $ equal to 11.1 and solve for $ x $:

\[
11.1 = 1.8208e^{0.3387x}
\]

**Steps to Solve:**
1. Divide both sides by 1.8208:
\[
\frac{11.1}{1.8208} = e^{0.3387x}
\]

2. Calculate the division:
\[
6.096 \approx e^{0.3387x}
\]

3. Take the natural logarithm (ln) of both sides to solve for $ x $:
\[
\ln(6.096) = 0.3387x
\]

4. Simplify using the properties of logarithms:
\[
x = \frac{\ln(6.096)}{0.3387}
\]

5. Compute the value of $ x $.

6. Add the computed value of $ x $ to the base year (1998) to find the approximate year.

You can use a calculator or a computational tool to find the exact value.

### Graphical Explanation:
If there were a graph depicting this function, it would show an exponential growth curve starting from the year 1998. The x-axis would represent the years, starting from 1998, and the y-axis would represent the export value in billions of dollars. The point where the curve intersects the horizontal line at $11.1 billion would indicate the year you need to determine.

### Additional Resources:
For further guidance, you can refer to the video by clicking on the “Question Help” link provided in the original material or use other educational resources available on exponential growth and logarithmic functions.

### Answer

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