How to prove that $\exp(x)$ and $\log(x)$ are inverse?

$\log(e^x)=\int_1^{e^x}\frac{1}{t}dt=\int_0^x\frac{1}{e^u}e^u du=x$ and since it's easy to prove that $e^x$ is bijective then $\log$ is its inverse.

We use the fact that $g(x)=\exp(x)$ is the unique function $g:\mathbb{R} \rightarrow (0,\infty)$ such that $g'(x)=g(x)$ and $g(0)=1.$

Since $f(x)=\log(x)$ is evidently bijective, it must have the inverse $f^{-1}:\mathbb{R} \rightarrow (0,\infty)$. By the well-known theorem on the derivative of the inverse function, $$(f^{-1}){'}(x)=\frac{1}{f'(f^{-1}(x))}=\frac{1}{1/(f^{-1}(x))}=f^{-1}(x)$$ for every $x>0$. Furthermore, since $f(1)=\log(1)=0,$ $f^{-1}(0)= 1$.

It follows that $f^{-1}(x)$ must be $\exp(x).$