Bacteria population grows proportionally; p(0)=1000, increases 20% in 2h.

Question

Bacteria population grows proportionally; $p(0)=1000$, increases 20\% in 2h. If population is 2000 at $t=\dfrac{k}{\log_e(6/5)}$h, then $\left(\dfrac{k}{\log_e2}\right)^2$ is:

Accepted Answer

$x=1000e^{\lambda t}$; at $t=2,x=1200$: $\lambda = \dfrac12\ln\dfrac65$. At $x=2000$, $t=k/\ln(6/5)$: $2=e^{k/2} \implies k=2\ln2$.

$$\left(\frac{k}{\ln2}\right)^2 = 4.$$

Solutions · 1