Answer
$R(t)=2.001(1.030)^t$
Work Step by Step
Given
$$
\begin{array}{c|c|c|c}
\hline t & 5 & 9 & 15 \\
\hline R(t) & 2.32 & 2.61 & 3.12 \\
\hline
\end{array}
$$
A function is linear if the rate of change is a constant.
$$
\begin{aligned}
&\frac{2.61-2.32}{9-5}=0.0725\\
&\frac{3.12-2.61}{15}=0.085 \text {, }
\end{aligned}
$$
The function is not linear. We now check the ratios to see if the function is exponential.
$$
\begin{aligned}
\frac{R(9)}{R(5)} & =\frac{a b^9}{a b^5}=\frac{2.61}{2.32} \\
\frac{b^9}{b^5} & =\frac{2.61}{2.32} \\
b^4 & =\frac{2.61}{2.32} \\
b & =\left(\frac{2.61}{2.32}\right)^{\frac{1}{4}} \approx 1.030
\end{aligned}
$$
$$
\begin{aligned}
\frac{R(9)}{R(5)}=\frac{a b^9}{a b^5} & =\frac{2.61}{2.32} \\
b^4 & =1.125 \\
b & =1.030 .
\end{aligned}
$$
We can safely assume that the function is approximately exponential. Let $R(t) = ab^t$,
Now:
$R(t)=a(1.030)^t$
Find $b$ from:
$$
\begin{aligned}
R(5) & =a(1.030)^5 \\
2.32 & =a(1.030)^5 \\
a & =\frac{2.32}{1.030^5} \approx 2.001
\end{aligned}
$$
$$
\text { Hence, } R(t)=2.001(1.030)^t
$$
