SOLUTION: Please help me solve this.... In September 1998 the population of the country of West Goma in millions was modeled by f(x)=17.5e^0.0006x. At the same time the population of East

Question 21493: Please help me solve this....
In September 1998 the population of the country of West Goma in millions was modeled by f(x)=17.5e^0.0006x. At the same time the population of East Goma in millions was modeled by g(x)=14.1e^0.0187x. In both formulas x is the year, where x=0 corresponds to September 1998. Assuming these trends continue, estimate the year when the population of West Goma will equal the population of East Goma.
The answer is 2010 but i have tried every method possible to come up with this answer please help......
Answer by mmm4444bot(95) (Show Source): You can put this solution on YOUR website!
Hello There:
To solve this problem algebraically, we need to know a property of logarithms.
ln(a*b) = ln(a) + ln(b)
In other words, if we take the natural log of a product, that is the same as adding the natural logs of the factors.
We need to find the value of x such that f(x) = g(x).
17.5*e^(0.0006*x) = 14.1*e^(0.0187*x)
Take the natural log of both sides.
ln[17.5*e^(0.0006*x)] = ln[14.1*e^(0.0187*x)]
Now use the property described above to write each side as a sum of natural logs.
Remember that ln(e^x) = x, so ln[e^(0.0006*x)] = 0.0006*x.
ln(17.5) + 0.0006*x = ln(14.1) + 0.0187*x
Group like terms; subtract 0.0006*x and ln(14.1) from both sides.
0.0181*x = ln(17.5) - ln(14.1)
Use a calculator to evaluate the expression on the right side.
0.0181*x = 0.2160
Solve for x.
x = 11.9
Adding 12 years to 1998 gives the desired result.
~ Mark

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