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Thread: How I can solve this question

  1. #1
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    How I can solve this question

    Hiu

    How I can solve this question
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by r-soy View Post
    Hiu

    How I can solve this question
    To graph the equation between 0 and 5 start with the ordered point (0,100)

    You know that in 1 hour the amount will be $\displaystyle 2^2 = 4$ times the amount at 0 so you get the second ordered pair of (1,400)

    Plot the others similarly and draw the graph

    Spoiler:
    Use the exponential growth equation $\displaystyle A_t = A_0e^{kt}$

    We know that $\displaystyle A_0 = 100$ and that in $\displaystyle t = \frac{1}{2}$ the number doubles - $\displaystyle A_t = 2A_0 = 200$

    Use this relationship to find the growth constant $\displaystyle k$

    $\displaystyle 200=100e^{\frac{1}{2}k}$

    $\displaystyle k = 2ln(2) = ln(4)$

    Spoiler:
    $\displaystyle A_t=A_0e^{t\,ln(4)} = A_0 \cdot 4^t$



    As the graph is exponential growth I suggest using a log scale on the y axis

    $\displaystyle ln(y) = kt + ln(A_0)$ which gives a straight line in the form $\displaystyle y=mx+c$


    Edit: Plot $\displaystyle ln(y)$ against $\displaystyle t$ to get the straight line, with gradient $\displaystyle k$ and y intercept $\displaystyle ln(A_0)$
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  3. #3
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    thanks e^(i*pi)
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  4. #4
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    There is no reason to use "e" or logarithms.

    There are 2N half hour periods in t hours so the population will have doubled 2t times: $\displaystyle N= (100) (2^{2t})$
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