Question

Use the formula P=le^kt. A bacterial culture has an initial population of 10,000. If its population declines to 7,000 in 4 hours, what will it be at the end of 6 hours?

Answer 1

$\bf \textit{Amount of Population Growth}\\\\ A=Ie^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &10000\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\\ \end{cases} \\\\\\ A=10000e^{rt}\\\\ -------------------------------$

$\bf \textit{population declines to 7,000 in 4 hours}\quad \begin{cases} A=7000\\ t=4 \end{cases} \\\\\\ 7000=10000e^{r4}\implies \cfrac{7000}{10000}=e^{4r}\implies \cfrac{7}{10}=e^{4r} \\\\\\ ln\left( \frac{7}{10} \right)=ln(e^{4r})\implies ln\left( \frac{7}{10} \right)=4r\implies \cfrac{ln\left( \frac{7}{10} \right)}{4}=r \\\\\\ -0.089\approx r\qquad thus\qquad \boxed{A=10000e^{-0.089t}}$

$\bf -------------------------------\\\\ \textit{what will it be at the end of 6 hours?}\quad t=6\implies A=10000e^{-0.089\cdot 6}$

and surely you know how much that is.