SOLUTION: The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208e^0.3387x, where x = 0 corresponds to 1998. In what year did/will the ex

Question 1166949: The value of China's exports of automobiles and parts (in billions of dollars) is approximately f(x)=1.8208e^0.3387x, where x = 0 corresponds to 1998. In what year did/will the exports reach $6.3 billion?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let y = f(x).
your equation becomes:
y = 1.8208*e^(.3387*x)
when y = 6.3, the equation becomes:
6.3 = 1.8208*e^(.3387*x)
divide both sides of this equation by 1.8208 to get:
6.3/1.8208 = e^(.3387*x)
take the natural log of both sides of this equation to get:
ln(6.3/1.8208) = ln(e^(.3387*x))
by the log property that says log(a^x) = x*log(a), this becomes:
ln(6.3/1.8208) = .3387*x*ln(e)
since ln(e) = 1, this becomes:
ln(6.3/1.8208) = .3387*x
solve for x to get:
x = ln(6.3/1.8208)/.3387 to get:
x = 3.664817444.
if you round this to 3 decimal places, it will be equal to 3.665.
3.665 is the same value that the desmos.com graph shows when the equation of y = 1.8208*e^(.3387*x) and y = 6.3 is the intersection of both equations on the same graph.
the intersection is at the point (3.665,6.3).
since that point is shown in (x,y) format, this means that y = 6.3 when x = 3.665.
the graph is shown below:

$$$

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