The functional form the given equation can be written as
\(\displaystyle{f{{\left({t}\right)}}}={2}{\left({1.08}\right)}^{{t}}\)
where f(0)=2 and 1.08 is the ratio of succesive values of the function.

That is, \(\displaystyle\frac{{f{{\left({1}\right)}}}}{{f{{\left({0}\right)}}}}=\frac{{f{{\left({2}\right)}}}}{{f{{\left({1}\right)}}}}=\ldots=\frac{{{f{{\left({t}+{1}\right)}}}}}{{f{{\left({t}\right)}}}}=\ldots={1.08}\) So, the recursive rule for this function van be written as \(\displaystyle{f{{\left({t}+{1}\right)}}}={1.08}\cdot{f{{\left({t}\right)}}}\)

That is, \(\displaystyle\frac{{f{{\left({1}\right)}}}}{{f{{\left({0}\right)}}}}=\frac{{f{{\left({2}\right)}}}}{{f{{\left({1}\right)}}}}=\ldots=\frac{{{f{{\left({t}+{1}\right)}}}}}{{f{{\left({t}\right)}}}}=\ldots={1.08}\) So, the recursive rule for this function van be written as \(\displaystyle{f{{\left({t}+{1}\right)}}}={1.08}\cdot{f{{\left({t}\right)}}}\)