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

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

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Liyana Mansell

E is a special number where ${e}^{x}$ differentiates to ${e}^{x}$. $\mathrm{ln}\left(x\right)$ 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 $fg\left(x\right)={e}^{g\left(x\right)}={e}^{\mathrm{log}x}=x$. $GF\left(x\right)=\mathrm{ln}\left(f\left(x\right)\right)=\mathrm{ln}\left({e}^{x}\right)=x\cdot \mathrm{ln}\left(e\right)$ (using log laws, bringing power to the front), $=x×1=x$.
Therefore $fg\left(x\right)=x=gf\left(x\right)$