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madison.evans
Group Title
PLEASE HELP ..
On January 1 2010, Chessville has a population growth. Its population increases 7% each year. On the same day checkersville has a population of 70,000 people. Cheakersveille starts to expeirence a population decline . It's population decreases 4% each year . During what year will the population of Chessville first exceed that of Checkersville?
 one year ago
 one year ago
madison.evans Group Title
PLEASE HELP .. On January 1 2010, Chessville has a population growth. Its population increases 7% each year. On the same day checkersville has a population of 70,000 people. Cheakersveille starts to expeirence a population decline . It's population decreases 4% each year . During what year will the population of Chessville first exceed that of Checkersville?
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
we are missing the initial population of "chessville"
 one year ago

madison.evans Group TitleBest ResponseYou've already chosen the best response.0
oh sorry its 50000
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[50000(1.07)^t=70,000(.96)^t\] solve for \(t\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
lets see, divide by \(50,000\) get \[(1.07)^t=1.4(.96)^t\] divide by \((.96)^t\) get \[\left(\frac{1.07}{.96}\right)^t=1.4\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
then use "change of base" formula to find \(t\) \[t=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
and of course a calculator
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
https://www.wolframalpha.com/input/?i=log%281.4%29%2Flog%281.07%2F.96%29
 one year ago
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