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 popultion decreases 4% each year. during what year will the population of chessville first exceed that of checkersville?
+33654 If n is the number of years for the populations to match then you want to solve:
50000*1.07n = 70000*0.96n for n
Take logs of both sides and use the fact that log(a*b) = log(a) + log(b) and log(an) = n*log(a)
log(50000) + n*log(1.07) = log(70000) + n*log(0.96)
log10(50000)+n×log10(1.07)=log10(70000)+n×log10(0.96)⇒n=(35885121×ln(70000)−35885121×ln(50000))3892841⇒n=3.1016799618306674
so n ≈ 3years.
+33654 If n is the number of years for the populations to match then you want to solve:
50000*1.07n = 70000*0.96n for n
Take logs of both sides and use the fact that log(a*b) = log(a) + log(b) and log(an) = n*log(a)
log(50000) + n*log(1.07) = log(70000) + n*log(0.96)
log10(50000)+n×log10(1.07)=log10(70000)+n×log10(0.96)⇒n=(35885121×ln(70000)−35885121×ln(50000))3892841⇒n=3.1016799618306674
so n ≈ 3years.
