pop(chessville) = 50,000(1.07)^n
pop(checkerville) = 70,000(.96)^n
we want to know when
50,000(1.07)^n ≥ 70,000(.96)^n
divide by 10,000
5(1.07)^n = 7(.96)^n
take logs of both sides and use rules of logs
log5 + nlog1.07 = log7 + nlog.96
nlog1.07 - nlog.96 = log7 - log5
n(log1.07 - log .96) = log7 - log5
n = (log7-log5)/(log1.07-log.96)
= appr 3.1 years
looks like they are equal after 3.1 years, so it will happen in the fourth year.
check:
after 3 years:
pop(chessville) = 50,000(1.07)^3 = 61,252
pop(checkerville) = 70,000(.96)^3 = 61,932
notice that Checkerville's population is still greater than Chessville's
On January 1, 2010, Chessville has a population of 50,000 people. Chessville then enters a period of population growth. Its population increases 7% each year. On the same day, Checkersville has a population of 70,000 people. Checkersville starts to experience a population decline. Its population decreases 4% each year. During what year will the population of Chessville first exceed that of Checkersville?
1st step: Interpret the concepts of the task and translate them into mathematics. (RESTATE THE PROBLEM IN YOUR OWN WORDS).
Stuck on the first step. What am I supposed to do?
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