Exponential Equations 18

Question

The value of China's exports of automobiles and parts (in billions of dollars) is approximately , where x = 0 corresponds to 1998.In what year did/will the exports reach $11.6 billion?

Answer 1

f(x) = 1.8208e^0.3387x

11.6 = 1.8208e^0.3387x

11.6/1.8208 = e^0.3387x

ln(11.6/1.8208) = 0.3387x

Use your calculator to solve for x. Since x is the number of years since 1998, add 1998 to x to get the year. Since the question just asks for the year, you can drop any decimals.

Answer 2

We have exponential equation: 1.8208e^0.3387x = 11.6; e^0.3387x = 11.6/1.8208; 0.3387x = ln(11.6/1.8208);

x = ln(11.6/1.8208)/0.3387 = 5.467≈ 5.5 years. So in 1998 + 5 = 2003 year this happened

Answer 3

Stephanie.

f(x) = 1.8208 e^{0.3387 x}

We want the exports to be 11.6 billion, in other words find x so that f(x) = 11.6

f(x) = 11.6

11.6 = 1.8208 e^{0.3387 x}

11.6 / 1.8208 = e^0.3387x

Taking logs to base e (ln) :

ln (11.6 / 1.8208) = ln (e ^0.3387x) = 0.3387x

So x = ln (11.6 / 1.8208) / 0.3387 = 5.467

x=0 corresponds to 1998

x=5 corresponds to 1998 + 5 = 2003

x=6 corresponds to 1998 + 6 = 2004

So as our calculated x is between 5 and 6, exports reach 11.6 billion during the year 2003.