[tex]\bf \qquad \textit{Amount for Exponential change}\\\\
A=P(1\pm r)^t\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{starting amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed period}\\
\end{cases}\\\\
-------------------------------\\\\[/tex]
[tex]\bf A=P(1+r)^t\qquad
\begin{cases}
\textit{hour 0, starting point}\\
t=0\qquad A=4000
\end{cases}\implies 182=ae^{k0}
\\\\\\
4000=P(1+r)^0\implies 4000=P
\\\\\\
thus\qquad A=4000(1+r)^t\\\\
-------------------------------\\\\[/tex]
[tex]\bf A=P(1+r)^t\qquad
\begin{cases}
\textit{8 hours later}\\
t=8\qquad A=4400
\end{cases}\implies 4400=4000(1+r)^8
\\\\\\
\cfrac{4400}{4000}=(1+r)^8\implies \cfrac{11}{10}=(1+r)^8\implies \sqrt[8]{\cfrac{11}{10}}=1+r
\\\\\\
\sqrt[8]{\cfrac{11}{10}}-1=r\implies 0.012\approx r\qquad thus\qquad \boxed{A=4000(1+0.012)^t}[/tex]
how many bacteria after 9 hours? well, 9hours later is just t = 9, plug that in, to get A
