Log word problem-college agebra

Question

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 $7.5 billion?

Answer 1

f(x) = 1.8208 e^{.3387x ;}   f(x) represents value of China's automobile exports

 

What year will f(x) = $7.5 billion ; 7,500,000,000

 

$7,500,000,000 = 1.8208 e^.3387x

 

Divide both sides of equation by 1.8208

 

   4119068541.3 = e^.3387x

 

take the ln of both sides

 

ln 4119068541.3 = ln e^.3387x

 

            22.1389 = (.3387x) ln e  ; (lne = 1)

 

            22.1389 = .3387x

 

 22.1389/ .3387 = x

 

             65.364 = x

 

Approximately 65 years from 1998 = the year 2063

 

 

Answer 2

7.5 billion = 7.5 x 10^9

7.5 x 10^9 = 1.8208e^(.3387x)

Divide each side by 1.8208

4.119 x 10^9 = e^(.3387x)

Take the natural log (Ln) of both sides.
Remember the identity that Ln (e^x) = x

Ln 4.119 x 10^9 = .3387x

use a calculator to compute:
Ln 4.119 x 10^9 = 22.139

22.139 = .3387x

Divide each side by .3387

x = 22.139/.3387 = 65.4

when x = 0, year = 1998
adding 65.4 to both sides gives:

when x = 65.4 year = 2063.4 which is in year 2063

China's exports will reach $7.5 billion dollars in the year 2063

Answer 3

It pays to check what units values are given in.  In this case the function represents billions of dollars.

f(x)=1.8208e^.3387x

 

When f(x) = 7.5 = 1.8208e^.3387x

 

Taking the log of both sides of this equation:

ln(7.5/1.8208) = .3387x

x = ln(7.5/1.8208)/.3387 ≈ 4.18 years

 

This will occur sometime in the year 2002.