SOLUTION: The population of town A is double that of town B, but it is decreasing at the rate of 5% per year. The population of town B is increasing at the rate of 5% per year. In how many y

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The population of town A is double that of town B, but it is decreasing at the rate of 5% per year. The population of town B is increasing at the rate of 5% per year. In how many y      Log On


   



Question 1084462: The population of town A is double that of town B, but it is decreasing at the rate of 5% per year. The population of town B is increasing at the rate of 5% per year. In how many years will the population of town B be double that of town A? At the end of this time, how will the population of town B compare with the initial population of town A?

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
Let b and 2b be the populations of towns B and A, respectively. The population of town A is decreasing by 5%, so it is 95% of what it was the year before. Likewise, town B's population is 105% of what it was the year before. So, when town B is twice as big as town A, we have:
2*(0.95)^n*2b=(1.05)^n*b [where n is the time, in years]
4*(0.95)^n=1.05^n
ln 4 +n ln(0.95)=n ln 1.05
ln 4=n(ln 1.05-ln 0.95)
n=13.85138344 years before town B is twice as big as town A. At that time, town B will be 1.965626988 times its original size of b, whereas town A started at 2b. ☺☺☺☺