SOLUTION: The doubling function (y = y base0 2^(1/D)) can be used to model exponential growth when the doubling time is D. The bacterium Escherichia coli has a doubling period of 0.32 h. A c
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-> SOLUTION: The doubling function (y = y base0 2^(1/D)) can be used to model exponential growth when the doubling time is D. The bacterium Escherichia coli has a doubling period of 0.32 h. A c
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Question 1058803: The doubling function (y = y base0 2^(1/D)) can be used to model exponential growth when the doubling time is D. The bacterium Escherichia coli has a doubling period of 0.32 h. A culture of E. coli starts with 100 bacteria.
a) Determine the equation for the number of bacteria, y, in x hours.
b) Graph your equation
c) Graph the inverse
d) Determine the equation of the inverse. What does this equation represent?
e) How many hours will it take for there to be 450 bacteria in the culture? Explain your strategy. Answer by josgarithmetic(39621) (Show Source):
Doubling Time is 0.32 hours;
Can you calculate k?
This is not a complete solution but maybe you have enough to continue.
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asked for more than just the above ...
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MODEL:
The exponential doubling model graphed:
The x-values of hours below 0 do not mean much.
The inverse would be TIME in hours as a function of POPULATION SIZE. If you do that, then y becomes time, and x becomes population size. Now, y represents an logarithmic function instead of a exponential function. The quick way to get to this is to take the logarithmic equation and switch the x and the y, and solve for y.
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