Answer:
This was occur sometimes in year 2002.4
Step-by-step explanation:
* Lets explain how to solve the problem
- The value of China's exports of automobiles and parts
(in billions of dollars) is approximately f(x) = 1.8208 e^(0.3387 x)
# You must pay attention about the function is already calculated in
billions dollars so you will not multiply the value of f(x) by 10^9 to
change it to billions
∵ The value of x = 0 at 1998
- Remember that e^(0) = 1
∴ f(0) = 1.8208 e^(0) = 1.8208 billion dollars
- You need to calculate the year that the export reaches 8 billion
∵ f(x) = 1.8208 e^(0.3387 x)
∵ f(x) = 8 billion
∴ 8 = 1.8208 e^(0.3387 x)
- Divide both sides by 1.8208
∴ 4.393673111 = e^(0.3387 x)
- Insert ㏑ in both sides
∴ ㏑(4.393673111) = ㏑[e^(0.3387 x)]
- Remember ㏑(e^n) = n ㏑(e), ㏑(e) = 1, then ㏑(e^n) = n
∴ ㏑(4.393673111) = 0.3387 x
- Divide both sides by 0.3387
∴ x = ㏑(4.393673111) ÷ 0.3387
∴ x = 4.37 ≅ 4.4
- Lets add the number of years to 1998
∴ The year is 1998 + 4.4 = 2002.4
∴ This was occur sometimes in year 2002.4
