91Ó°ÊÓ

A culture of bacteria obeys the law of uninhibited growth.

If N is the number of bacteria in the culture and t is the time in hours, then N as a function of t can be expressed as$N\left(t\right)=N_0{e}^{kt}$

Substitute $N\left(t\right)=800,N_0=500,t=1$in the function $N\left(t\right)=N_0{e}^{kt}$

$$\begin{gathered} N\left(t\right)=N_0{e}^{kt} \\ 800=500e^{k×1} \\ \frac{8}{5}=e^k \\ \ln \left(\frac{8}{5}\right)=\ln \left(e^k\right) \\ \ln \left(\frac{8}{5}\right)=k \\ k=0.47 \end{gathered}$$

Now bacteria present in the culture after 5 hours can be calculated by

substituting role="math" localid="1647285647109" $N_0=500,k=0.47,t=5$in the function $N\left(t\right)=N_0{e}^{kt}$

role="math" localid="1647285825350" $$\begin{gathered} N\left(t\right)=N_0{e}^{kt} \\ N\left(5\right)=500e^{0.47×5} \\ N\left(5\right)≈5243 \end{gathered}$$

Bacteria in the culture after 5 hours will be 5243

Substitute $N\left(t\right)=20000,N_0=500,k=0.47$in the function $N\left(t\right)=N_0{e}^{kt}$

$$\begin{gathered} N\left(t\right)=N_0{e}^{kt} \\ 20000=500e^{0.47t} \\ 40=e^{0.47t} \\ \ln \left(40\right)=\ln \left(e^{0.47t}\right) \\ \ln \left(40\right)=0.47t \\ 3.68=0.47t \\ t=\frac{3.68}{0.47} \\ t=7.82 \end{gathered}$$

It will take 7.82 hours for the bacteria to be 20,000