Question

During its exponential growth phase a colony of the bacteria E. coli grew from 5052 cells to 39,845 cells in 60 minutes how long will it take the colony of 5052 cells to grow to 200,000 cells?

Answer 1

at minute 0, or t=0, the colony had an amount of 5052, let's use that firstly,


`\bf \textit{Amount of Population Growth}\\\\ A=Ie^(rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &5052\\ I=\textit{initial amount}\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\to &0\\ \end{cases} \\\\\\ 5052=Ie^(r0)\implies 5052=I\cdot 1\implies 5052=I\qquad thus \\\\\\ A=5052e^(rt)\\\\ -------------------------------\\\\`


`\bf \textit{now, after 60 minutes, t = 60, they'd grown to 39845} \\\\\\ 39845=5052e^(r60)\implies \cfrac{39845}{5052}=e^(r60)\implies ln\left( (39845)/(5052) \right)=ln(e^(60r)) \\\\\\ ln\left( (39845)/(5052) \right)=60r\implies \cfrac{ln\left( (39845)/(5052) \right)}{60}=r\implies \boxed{0.0344\approx r} \\\\\\ thus\qquad A=5052e^(0.0344t)`

now, how long will it take it to become 200,000?


`\bf A=5052e^(0.0344t)\implies 200000=5052e^(0.0344t) \\\\\\ \cfrac{200000}{5052}=e^(0.0344t)\implies ln\left( (200000)/(5052) \right)=ln(e^(0.0344t)) \\\\\\ ln\left( (200000)/(5052) \right)=0.0344t \implies \cfrac{ln\left( (200000)/(5052) \right)}{0.0344}=t\impliedby minutes`