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Math Help - constant interest

  1. #1
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    Exclamation constant interest

    In 1990 the population of a city was about 200,000. If the population increases at a constant rate of 1.5% per year, in what year is the population projected to reach 250,000?

    Please help im going crazy!!
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  2. #2
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    Hello,

    Consider the function x(t) the number of inhabitants and t the number of years.

    For the first year, 200000 becomes 200000*(100+1.5)/100
    For the second year, the population becomes 200000*(100+1.5)/100 *(100+1.5)/100 = 200000*((100+1.5)/100)

    And so on... Each year, you multiply the number of inhabitants by (100+1.5)/100 (corresponds to an augmentation of 1.5%)

    So x(t)=200000*(101.5/100)^t

    And you want t such as x(t)>250,000

    Hence the problem is to solve 200000*(101.5/100)^t > 250000
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    i still dont get this...
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  4. #4
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    Well,

    The initial population is at N=200,000 inhabitants

    The year following, it increases of 1.5%, which means that to N, we add 1.5% of N
    The population will then be N1= N+\frac{1.5}{100} N = N(\frac{100}{100}+\frac{1.5}{100}) = N(\frac{101.5}{100})

    The second year, it increases again of 1.5%, which means that to N1, we add 1.5% of N1
    With the same scheme, we got N2= N1(\frac{101.5}{100}) = N(\frac{101.5}{100})^2

    So the general expression for the population at year n is :

    N (\frac{101.5}{100})^n


    And you want to know n such as the population at this year will reach 250,000 (or more).

    So solve 200,000 * (\frac{101.5}{100})^n > 250,000

    If we divide both sides by 200,000 , we have :

    (\frac{101.5}{100})^n > \frac{250000}{200000} = \frac{5}{4}

    Then use the logarithm :

    n \ln(\frac{101.5}{100}) > \ln(\frac{5}{4}) = 0.223

    \ln(\frac{101.5}{100}) = 0.0149

    So n > \frac{0.223}{0.0149}

    n > 14.9

    As the number of years is an integer, the population will reach 250,000 within 15 years.
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  5. #5
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    Hello, imstartingtohatethis!

    In 1990 the population of a city was about 200,000.
    If the population increases at a constant rate of 1.5% per year,
    in what year is the population projected to reach 250,000?
    You're expected to know the Exponential Growth formula.

    For this problem it is: . P \;=\;P_o(1 + r)^n
    . . where P_o is the initial population,
    . . n is the number of years since 1990,
    . . and r is the rate of increase.

    So we have: . P \;=\;200,000(1.015)^n


    When does P = 250,000 ?

    We have: . 200,000(1.015)^n \;=\;250,000

    Divide by 20,000: . 1.015^n \:=\:1.25

    Take logs: . \ln(1.015^n) \;=\;\ln(1.25) \quad\Rightarrow\quad n\!\cdot\!\ln(1.015) \;=\;\ln(1.25)

    Therefore: . n \;=\;\frac{\ln(1.25)}{\ln(1.015)} \;=\;14.98753167 \;\approx\;15


    The population will reach 250,000 in about 15 years . . . in 2005.

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