Question

A population of bacteria is growing according to the exponential model P = Ce^kt, where P is the number of colonies and t is measured in hours. If there were 120 colonies present initially, and there are 550 colonies present after 4 hours, what is k, the population's growth rate?

Answer 1

we know that initially there were 120 colonies or namely that C = 120, well, what time was that? that was the hour 0, or namely t = 0, so then

$P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\\ C=\textit{initial amount}\dotfill &120\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &0\\ \end{cases} \\\\\\ P=120e^{r\cdot 0}\implies P=120e^0\implies P=120\cdot 1\implies P=120$

we also know that when t = 4, C = 550

$\textit{Amount of Population Growth} \\\\ P=Ce^{rt}\qquad \begin{cases} P=\textit{accumulated amount}\dotfill&120\\ C=\textit{initial amount}\dotfill &550\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &4\\ \end{cases} \\\\\\ 120=550e^{4r}\implies \cfrac{120}{550}=e^{4r}\implies \cfrac{12}{55}=e^{4r}~\hfill \stackrel{\textit{using this log cancellation rule}}{\log_a(a^x)=x}$

$\log_e\left( \frac{12}{55}\right)=\log_e(e^{4r})\implies \log_e\left( \frac{12}{55}\right)=4r\implies \ln\left( \frac{12}{55}\right)=4r \\\\\\ \cfrac{\ln\left( \frac{12}{55}\right)}{4}=r\implies -0.381\approx r = k$