Question

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?

Answer 1

`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`