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