as "t" increases, notice, the P is also increasing from its previous value, since that's the case, then is a "growth" equation,
[tex]\bf \qquad \textit{Amount for Exponential Growth}\\\\
P=I(r)^t\qquad
\begin{cases} P=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\\
t=\textit{elapsed time}\\
\end{cases}\\\\
-------------------------------\\\\
\textit{we know that }
\begin{cases}
t=0\\
P=4
\end{cases}\implies 4=I(r)^0\implies 4=I\cdot 1\implies 4=I
\\\\\\
therefore\qquad P=4(r)^t\\\\
-------------------------------\\\\[/tex]
[tex]\bf \textit{we also know that }
\begin{cases}
t=1\\
P=6
\end{cases}\implies 6=4(a)^1\implies \cfrac{6}{4}=a^1
\\\\\\
\cfrac{3}{2}=a\qquad therefore\qquad \boxed{P=4\left(\frac{3}{2} \right)^t}\\\\
-------------------------------\\\\
\textit{now, when t = 3 and t = 4, what is \underline{P}?}
\\\\\\
P=4\left(\frac{3}{2} \right)^3\qquad \qquad \qquad P=4\left(\frac{3}{2} \right)^4[/tex]
