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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?

Respuesta :

[tex]\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}\\\\ -------------------------------\\\\[/tex]

[tex]\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}}[/tex]

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

and surely you know how much that is.
abidemiokin

The required population after 6 hours will be 28,576 populations

Given the formula

[tex]P=le^{kt}[/tex] where:

I is the initial population = 10,000

P is the population equivalent after 4 hours

t is the time taken = 4hours

Substitute the given values into the formula

[tex]7000=1000e^{4k}\\\frac{7000}{10000}=e^{4k}\\0.7=e^{4k}\\ln0.7=lne^{4k}\\0.7=4k\\k=\frac{0.7}{4}\\k= 0.175[/tex]

Next is to get the population size after 6 hours

[tex]P=10000e^{6(0.175)}\\\\P=10000e^{1.05}\\P=10000(2.8576)\\P=28576[/tex]

Hence the required population after 6 hours will be 28,576 populations

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