A colony of bacteria is grown under ideal conditions in a lab so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. at the end of 5 hours there are 40000. How many bacteria were present initially?

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A colony of bacteria is grown under ideal conditions in a lab so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. at the end of 5 hours there are 40000. How many bacteria were present initially?

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Answer 1 · 2016-05-16T14:00:32

$6454$

Explanation:

The law of exponential grow is written as
$y = Y_0 e^(alpha t)$ where $Y_0$ is the initial population
$alpha$ the coefficient of exponential grow and $t$ time.
When $t=3 times 60 times 60$ [s] the population is $10000$
so $10000 = Y_0 e^(alpha 4800)$
and for $t = 5 times 60 times 60$ [s] we have
$40000=Y_0 e^(alpha 18000)$
Dividing side by side both equations
$40000/10000 = e^(alpha 15200)->alpha_0 = 0.0000912036$
Now $Y_0=10000/e^(alpha_0 4800) approx 6454$

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A colony of bacteria is grown under ideal conditions in a lab so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. at the end of 5 hours there are 40000. How many bacteria were present initially?

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