To determine the year in which China's exports of automobiles and parts reached [tex]$6.1$[/tex] billion dollars, we start by using the given function [tex]\( f(x) = 1.8208 \cdot e^{0.3387x} \)[/tex], where [tex]\( x \)[/tex] is the number of years since 1998.
We are asked to find the value of [tex]\( x \)[/tex] when [tex]\( f(x) = 6.1 \)[/tex]. Thus, we can set up the equation:
[tex]\[
6.1 = 1.8208 \cdot e^{0.3387x}
\][/tex]
First, isolate the exponential term by dividing both sides of the equation by 1.8208:
[tex]\[
\frac{6.1}{1.8208} = e^{0.3387x}
\][/tex]
Next, compute the value of the left side:
[tex]\[
\frac{6.1}{1.8208} \approx 3.3493
\][/tex]
Now, we have:
[tex]\[
3.3493 = e^{0.3387x}
\][/tex]
To solve for [tex]\( x \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[
\ln(3.3493) = \ln(e^{0.3387x})
\][/tex]
Using the property of logarithms that [tex]\(\ln(e^a) = a\)[/tex], this simplifies to:
[tex]\[
\ln(3.3493) = 0.3387x
\][/tex]
Next, solve for [tex]\( x \)[/tex] by dividing both sides by 0.3387:
[tex]\[
x = \frac{\ln(3.3493)}{0.3387} \approx 3.5696
\][/tex]
Since [tex]\( x \)[/tex] represents the number of years after 1998, the year when the exports reach [tex]$6.1 billion dollars is given by:
\[
1998 + x = 1998 + 3.5696 \approx 2001.5696
\]
To find the nearest year:
\[
2001.5696 \approx 2002
\]
Thus, China's exports of automobiles and parts reached $[/tex]6.1 billion dollars around the year [tex]\(\boxed{2002}\)[/tex].
