Answer
$F(t)=50,000(1.5)^{t/2}$,
where $t=0$ is two years ago.
Work Step by Step
Let $t=0$ represent two years ago.
Let $F(t)$ be the number of frogs, $t$ years from the beginning.
We want an exponential model, $F(t)=Ab^{t}$
The initial value at $t=0 $ is $A=50,000$
$(2,75000)$ - now - $\Rightarrow\left\{\begin{array}{l}
75,000=50,000 b^{2}\\
b^{2}=75000/50000\\
b^{2}=3/2\\
b=\sqrt{1.5}
\end{array}\right.$
$F(t)=50,000(\sqrt{1.5})^{t}$
$F(t)=50,000(1.5^{1/2})^{t}$
$F(t)=50,000(1.5)^{t/2}$
