The continuous growth rate model stipulates that the rate of population change [tex]\dfrac{\mathrm dP}{\mathrm dt}[/tex] is in proportion with the actual population [tex]P(t)[/tex] (a function of time [tex]t[/tex]). That is, there is some [tex]k[/tex] such that
[tex]\dfrac{\mathrm dP}{\mathrm dt}=kP[/tex]
Solving for [tex]P[/tex], you get
[tex]\dfrac{\mathrm dP}{P}=k\,\mathrm dt\implies \ln|P|=kt+C\implies P=Ce^{kt}[/tex]
You're given that [tex]k=1.38>0[/tex], so the population is increasing. At time [tex]t=0[/tex], you start with 2 bacteria, so [tex]P(0)=2[/tex] and
[tex]2=Ce^{k\times0}\implies C=2[/tex]
So, the growth equation is
[tex]P(t)=2e^{1.38t}[/tex]
