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Math Help - doubling bacteria

  1. #1
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    doubling bacteria

    A bacteria culture starts with 800 bacteria and doubles in size every half hour.

    I have already found the formula to be:
    f(x)=800(2)^{2t}

    I just am having trouble doing this question...

    (d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)
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  2. #2
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    Quote Originally Posted by ryan18 View Post
    A bacteria culture starts with 800 bacteria and doubles in size every half hour.

    I have already found the formula to be:
    f(x)=800(2)^{2t}

    I just am having trouble doing this question...

    (d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)

    Or should it be f(x)=800(2)^{t} ?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ryan18 View Post
    A bacteria culture starts with 800 bacteria and doubles in size every half hour.

    I have already found the formula to be:
    f(x)=800(2)^{2t}

    I just am having trouble doing this question...

    (d) Graph the population function and estimate the time for the population to reach 70,000. (Give your answer to the nearest hundredth.)
    this is just an a^x-graph, where a is a constant. they have a general form, they look like the e^x-graph which you should know. the difference here is the y-intercept is 800.

    for the estimate, plug in 70,000 for f(x) and solve for t. round to the nearest hundredth
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  4. #4
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    I dont know what to do once I get it down to

    87.5 = 2^{2t}

    Do I take the Log_2 of each side to get 2t? How do I compute that for the 87.5?
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  5. #5
    Senior Member Stroodle's Avatar
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    \frac{175}{2}=2^{2t}

    log_2\left ( \frac{175}{2}\right )=2t

    t=\frac{1}{2}log_2\left ( \frac{175}{2}\right )

    t\approx 3.23
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ryan18 View Post
    I dont know what to do once I get it down to

    87.5 = 2^{2t}

    Do I take the Log_2 of each side to get 2t? How do I compute that for the 87.5?
    you can use the change of base formula: remember, \log_a b = \frac {\log_c b}{\log_c a}

    so, \log_2 x = \frac {\log x}{\log 2}, and you can find \log on your calculator.
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  7. #7
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     87.5 = 2^{2t}

     \log_287.5 = 2t

     t = \frac{\log_287.5}{2}

    Use calculator on RHS.
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