# 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?

May 16, 2016

$6454$

#### 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
$\frac{40000}{10000} = {e}^{\alpha 15200} \to {\alpha}_{0} = 0.0000912036$
Now ${Y}_{0} = \frac{10000}{e} ^ \left({\alpha}_{0} 4800\right) \approx 6454$