Calibrate exponential bacterial growth

Let $x(t)$ count bacteria, taking $t=0$ at the first observation and measuring $t$ in hours. Then $dx/dt=kx$ with $k>0$. Separation gives $\ln x=kt+C$, so $x=Ae^{kt}$. The initial value gives $A=10^6$.

$x(1)=2\times10^6$ gives $e^k=2$, so $k=\ln2$ and $x(t)=10^6\,2^t$. At $t=4$, $x=16\times10^6=1.6\times10^7$. This follows the printed solution in measuring four hours from the first observation.