# A colony of bacteria originally contains 400 bacteria. It doubles in size every 30 minutes. How many hours will it take for the colony to contain 4,000 bacteria?

A colony of bacteria originally contains 400 bacteria. It doubles in size every 30 minutes. How many hours will it take for the colony to contain 4,000 bacteria?
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Coleman Ali
Recall the following population growth formula:
$P={P}_{0}{e}^{rt}$
Here, P reresents the total population after time t, ${P}_{0}$ represents the initial population and r represents the rate of growth.
Given,

Substitute the given values in the above-mentioned formula as,
$2{P}_{0}={P}_{0}{e}^{\frac{1}{2}r}\phantom{\rule{0ex}{0ex}}2={e}^{\frac{r}{2}}\phantom{\rule{0ex}{0ex}}\mathrm{ln}2=\frac{r}{2}\mathrm{ln}e\phantom{\rule{0ex}{0ex}}r=2\mathrm{ln}2$
This gives $r=2\mathrm{ln}2$
Now, calculate the required time when population reaches 4000 as,
$4000=\left(400\right){e}^{2\left(\mathrm{ln}2\right)t}\phantom{\rule{0ex}{0ex}}{e}^{\left(2\mathrm{ln}2\right)t}=\frac{4000}{400}\phantom{\rule{0ex}{0ex}}{e}^{\left(2\mathrm{ln}2\right)t}=10\phantom{\rule{0ex}{0ex}}ln{e}^{\left(2\mathrm{ln}2\right)t}=\mathrm{ln}10\phantom{\rule{0ex}{0ex}}\left(2\mathrm{ln}2\right)t=\mathrm{ln}10\phantom{\rule{0ex}{0ex}}t=\frac{\mathrm{ln}10}{2\mathrm{ln}2}\phantom{\rule{0ex}{0ex}}=1.7$
Therefore, it takes 1.7 hours for the colony to contain 4000 bacteria.