# If f(x) = e^x and g(x) = log x show that fog=gof given x>0

Question
Functions
If $$\displaystyle{f{{\left({x}\right)}}}={e}^{{x}}$$ and g(x) = log x show that fog=gof given x>0

2021-01-14
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)

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