In this problem, we have a continuous exponential growth model
so
the equation is of the form
[tex]y=a(e)^{kt}[/tex]where
a is the initial value ------> a=2,100
y is the number of bacteria
x ----> number of hours
so
[tex]y=2,100(e)^{kt}[/tex]For x=2.5 hours, y=2,249 bacteria
substitute
[tex]2,249=2,100(e)^{(2.5k)}[/tex]solve for k
apply ln both sides
[tex]\ln (\frac{2,249}{2,100})=2.5k\cdot\ln (e)[/tex]k=0.0274
convert to percentage
