Respuesta :
at minute 0, or t=0, the colony had an amount of 5052, let's use that firstly,
[tex]\bf \textit{Amount of Population Growth}\\\\ A=Ie^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\to &5052\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &0\\ \end{cases} \\\\\\ 5052=Ie^{r0}\implies 5052=I\cdot 1\implies 5052=I\qquad thus \\\\\\ A=5052e^{rt}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{now, after 60 minutes, t = 60, they'd grown to 39845} \\\\\\ 39845=5052e^{r60}\implies \cfrac{39845}{5052}=e^{r60}\implies ln\left( \frac{39845}{5052} \right)=ln(e^{60r}) \\\\\\ ln\left( \frac{39845}{5052} \right)=60r\implies \cfrac{ln\left( \frac{39845}{5052} \right)}{60}=r\implies \boxed{0.0344\approx r} \\\\\\ thus\qquad A=5052e^{0.0344t}[/tex]
now, how long will it take it to become 200,000?
[tex]\bf A=5052e^{0.0344t}\implies 200000=5052e^{0.0344t} \\\\\\ \cfrac{200000}{5052}=e^{0.0344t}\implies ln\left( \frac{200000}{5052} \right)=ln(e^{0.0344t}) \\\\\\ ln\left( \frac{200000}{5052} \right)=0.0344t \implies \cfrac{ln\left( \frac{200000}{5052} \right)}{0.0344}=t\impliedby minutes[/tex]
[tex]\bf \textit{Amount of Population Growth}\\\\ A=Ie^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\to &5052\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &0\\ \end{cases} \\\\\\ 5052=Ie^{r0}\implies 5052=I\cdot 1\implies 5052=I\qquad thus \\\\\\ A=5052e^{rt}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \textit{now, after 60 minutes, t = 60, they'd grown to 39845} \\\\\\ 39845=5052e^{r60}\implies \cfrac{39845}{5052}=e^{r60}\implies ln\left( \frac{39845}{5052} \right)=ln(e^{60r}) \\\\\\ ln\left( \frac{39845}{5052} \right)=60r\implies \cfrac{ln\left( \frac{39845}{5052} \right)}{60}=r\implies \boxed{0.0344\approx r} \\\\\\ thus\qquad A=5052e^{0.0344t}[/tex]
now, how long will it take it to become 200,000?
[tex]\bf A=5052e^{0.0344t}\implies 200000=5052e^{0.0344t} \\\\\\ \cfrac{200000}{5052}=e^{0.0344t}\implies ln\left( \frac{200000}{5052} \right)=ln(e^{0.0344t}) \\\\\\ ln\left( \frac{200000}{5052} \right)=0.0344t \implies \cfrac{ln\left( \frac{200000}{5052} \right)}{0.0344}=t\impliedby minutes[/tex]
