The value of china's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208e.^3387x, where x = 0 corresponds to 1998.
in what year did/will the exports reach $8 billion?
give your answer as the year, with at least one decimal place

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

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,