question 7\n0/5 pts 2 98 details\nthe value of chinas exports of automobiles and parts (in billions of…

question 7\n0/5 pts 2 98 details\nthe value of chinas exports of automobiles and parts (in billions of dollars) is approximately $f(x)=1.8208e^{.3387x}$, where $x = 0$ corresponds to the beginning of 2019.\nin what year did/will the exports reach $11.5 billion?\nquestion help: video post to forum\nsubmit question\nquestion 8

question 7\n0/5 pts 2 98 details\nthe value of chinas exports of automobiles and parts (in billions of dollars) is approximately $f(x)=1.8208e^{.3387x}$, where $x = 0$ corresponds to the beginning of 2019.\nin what year did/will the exports reach $11.5 billion?\nquestion help: video post to forum\nsubmit question\nquestion 8

Answer

Explanation:

Step1: Set up the equation

Set ( f(x)=11.5 ), so ( 11.5 = 1.8208e^{0.3387x} ).

Step2: Solve for ( e^{0.3387x} )

Divide both sides by ( 1.8208 ): ( \frac{11.5}{1.8208}=e^{0.3387x} ). Calculate ( \frac{11.5}{1.8208}\approx6.316 ), so ( 6.316 = e^{0.3387x} ).

Step3: Take the natural - logarithm of both sides

Using the property ( \ln(e^{a}) = a ), we have ( \ln(6.316)=0.3387x ). Since ( \ln(6.316)\approx1.843 ), then ( 1.843 = 0.3387x ).

Step4: Solve for ( x )

Divide both sides by ( 0.3387 ): ( x=\frac{1.843}{0.3387}\approx5.44 ).

Answer:

Since ( x = 0 ) corresponds to the beginning of 2019, and ( x\approx5.44 ), the year is ( 2019 + 5=2024 ).