E is a special number where \(\displaystyle{e}^{{x}}\) differentiates to \(\displaystyle{e}^{{x}}\). \(\displaystyle{\ln{{\left({x}\right)}}}\) is log base e of x.

Given that log base 10 of 10 is 1, and log base n of n is 1, ln(e) is also 1. So \(\displaystyle{f}{g{{\left({x}\right)}}}={e}^{{{g{{\left({x}\right)}}}}}={e}^{{{\log{{x}}}}}={x}\). \(\displaystyle{G}{F}{\left({x}\right)}={\ln{{\left({f{{\left({x}\right)}}}\right)}}}={\ln{{\left({e}^{{x}}\right)}}}={x}\cdot{\ln{{\left({e}\right)}}}\) (using log laws, bringing power to the front), = x*1 = x.

Therefore fg(x) = x = gf(x)

Given that log base 10 of 10 is 1, and log base n of n is 1, ln(e) is also 1. So \(\displaystyle{f}{g{{\left({x}\right)}}}={e}^{{{g{{\left({x}\right)}}}}}={e}^{{{\log{{x}}}}}={x}\). \(\displaystyle{G}{F}{\left({x}\right)}={\ln{{\left({f{{\left({x}\right)}}}\right)}}}={\ln{{\left({e}^{{x}}\right)}}}={x}\cdot{\ln{{\left({e}\right)}}}\) (using log laws, bringing power to the front), = x*1 = x.

Therefore fg(x) = x = gf(x)