To find f(x):

From the table it is clear that for every increment of x by 1 unit, the value of f(x) gets multiplied by 2.

So, let \(\displaystyle{f{{\left({x}\right)}}}={k}{.2}^{{x}}\).

Given \(\displaystyle{f{{\left({0}\right)}}}={4.5}\Rightarrow{k}{.2}^{{0}}={4.5}\Rightarrow{k}={4.5}\)

Hence \(\displaystyle{f{{\left({x}\right)}}}={\left({4.5}\right)}{.2}^{{x}}\)

To find g(x):

From the table it is clear that for every increment of x by 1 unit, the value of f(x) get halved.

So, let \(\displaystyle{g{{\left({x}\right)}}}={k}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}\)

Given: \(\displaystyle{g{{\left({0}\right)}}}={4}\Rightarrow{k}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{0}}={4}\Rightarrow{k}={4}\)

Hence \(\displaystyle{g{{\left({x}\right)}}}={4}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}\)

Thus, the exponential modes of two function are \(\displaystyle{f{{\left({x}\right)}}}={\left({4.5}\right)}{.2}^{{x}},\ {g{{\left({x}\right)}}}={4}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}\)