SOLUTION: Hi, I need help with this please. The value of China’s exports of automobiles and parts (in billions of dollars) is approximately f(x) =1.8208_e^0.3387x, where x=0 corresponds

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hi, I need help with this please. The value of China’s exports of automobiles and parts (in billions of dollars) is approximately f(x) =1.8208_e^0.3387x, where x=0 corresponds       Log On


   

Question 1198693: Hi, I need help with this please.
The value of China’s exports of automobiles and parts (in billions of dollars) is approximately f(x) =1.8208_e^0.3387x, where x=0 corresponds to 1998. In what year did/will the exports reach $6.4 billion?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29+=1.8208%2Ae%5E%280.3387x%29, where x=0 corresponds to 1998.
In what year did/will the exports reach $6.4 billion?
f%28x%29+=6.4
6.4=1.8208%2Ae%5E%280.3387x%29..........take natural log of both sides
ln%286.4%29=ln%281.8208%2Ae%5E%280.3387x%29%29
ln%286.4%29=ln%281.8208%29%2Bln%28e%5E%280.3387x%29%29
ln%286.4%29-ln%281.8208%29=%280.3387x%29ln%28e%29...........ln%28e%29=1
ln%286.4%29-ln%281.8208%29=0.3387x
x=%28ln%286.4%29-ln%281.8208%29%29%2F0.3387
x=3.71131years
x=3 years 8 months 16 days
if x=0 corresponds to 1998, then
x=3.71131 corresponds to 1998%2B3years 8 months 16 days
2001year 8 months 16 days
the exports will reach $6.4 billion in 2001