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madison.evans

  • 3 years ago

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?

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  1. anonymous
    • 3 years ago
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    we are missing the initial population of "chessville"

  2. madison.evans
    • 3 years ago
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    oh sorry its 50000

  3. anonymous
    • 3 years ago
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    \[50000(1.07)^t=70,000(.96)^t\] solve for \(t\)

  4. anonymous
    • 3 years ago
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    lets see, divide by \(50,000\) get \[(1.07)^t=1.4(.96)^t\] divide by \((.96)^t\) get \[\left(\frac{1.07}{.96}\right)^t=1.4\]

  5. anonymous
    • 3 years ago
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    then use "change of base" formula to find \(t\) \[t=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]

  6. anonymous
    • 3 years ago
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    and of course a calculator

  7. anonymous
    • 3 years ago
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    https://www.wolframalpha.com/input/?i=log%281.4%29%2Flog%281.07%2F.96%29

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