The general exponential growth model is defined as
[tex]y(t)=y_0e^{kt}[/tex]
Here ley y be the population of a bacteria at ttime in hours, k is the growth parameter, t is the time, y0 is the initial population of the bacteria.
Accoridng to this problem the initial population is 2300.
So,
[tex]y_0=2300.[/tex][tex]y(t)=2300e^{kt}[/tex]
It is given that after 2 hours, the population of the bacetria is 2450.
That is, at time t=2,y=2450.
[tex]\begin{gathered} y(t)=2300e^{kt} \\ y(2)=2300e^{k\cdot2} \\ 2450=2300e^{2k} \\ \frac{2450}{2300}=e^{2k} \\ \frac{49}{46}=e^{2k} \\ \ln (\frac{49}{46})=\ln (e^{2k}) \\ \ln (\frac{49}{46})=2k \\ \frac{1}{2}\ln (\frac{49}{46})=k \\ 0.031582=k \end{gathered}[/tex]
Now to get the percentage multiply the obatined k value by 100.
[tex]\begin{gathered} k=0.031582\cdot100 \\ =3.1582 \\ =3.16 \end{gathered}[/tex]
So, the required growth rate is k=3.16%.
