Questions: The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208 e^.3387 x, where x=0 corresponds to 1998. In what year did/will the exports reach 9.6 billion?

The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208 e^.3387 x, where x=0 corresponds to 1998.

In what year did/will the exports reach 9.6 billion?

Transcript text: The value of China's exports of automobiles and parts (in billions of dollars) is approximately $f(x)=1.8208 e^{.3387 x}$, where $x=0$ corresponds to 1998. In what year did/will the exports reach $\$ 9.6$ billion? $\square$

Solution

Solution Steps

Step 1: Start with the equation $A e^{kx} = B$

Step 2: Divide both sides by $A$ to get $e^{kx} = \frac{B}{A}$

Step 3: Take the natural logarithm ($\ln$) of both sides to get $kx = \ln(\frac{B}{A})$

Step 4: Solve for $x$ to find $x = \frac{1}{k} \ln(\frac{B}{A})$

Since the base year is 1998, the corresponding year is 1998 + 5 = 2003

Final Answer: $x = 5$

Corresponding Year: 2003

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Questions: The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208 e^.3387 x, where x=0 corresponds to 1998. In what year did/will the exports reach 9.6 billion?

Solution

Solution Steps

Final Answer: \(x = 5\)

Questions asked by other users