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

$6454$ $6454$

Explanation:

The law of exponential grow is written as
$y = Y_{0} e^{α t}$ $y = Y_0 e^(alpha t)$ where $Y_{0}$ $Y_0$ is the initial population
$α$ $alpha$ the coefficient of exponential grow and $t$ $t$ time.
When $t = 3 \times 60 \times 60$ $t=3 times 60 times 60$ [s] the population is $10000$ $10000$
so $10000 = Y_{0} e^{α 4800}$ $10000 = Y_0 e^(alpha 4800)$
and for $t = 5 \times 60 \times 60$ $t = 5 times 60 times 60$ [s] we have
$40000 = Y_{0} e^{α 18000}$ $40000=Y_0 e^(alpha 18000)$
Dividing side by side both equations
$\frac{40000}{10000} = e^{α 15200} \to α_{0} = 0.0000912036$ $40000/10000 = e^(alpha 15200)->alpha_0 = 0.0000912036$
Now $Y_{0} = \frac{10000}{e_{0}^{α_{0} 4800}} \approx 6454$ $Y_0=10000/e^(alpha_0 4800) approx 6454$

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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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