IDNLearn.com offers a user-friendly platform for finding and sharing knowledge. Join our interactive Q&A community and get reliable, detailed answers from experienced professionals across a variety of topics.
Sagot :
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
[tex]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[/tex]
we also know that when t = 4, C = 550
[tex]\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}[/tex]
[tex]\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[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
