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If we take the definitions of $\exp$ and $\ln$ as follows:

  • $\exp(x) = {\large\sum\limits_{i=0}^\infty} \dfrac{x^i}{i!}$
  • $\ln(x) = {\large\int_1^x} \dfrac1t\ dt$

how could we prove that these functions are inverses?

Neither $$\exp(\ln(x)) = \sum^\infty_{i=0}\frac{\left(\int_1^x\frac 1t\ dt\right)^i}{i!}$$ nor $$\ln(\exp(x)) = \int_1^{\sum\limits_{i=0}^\infty \frac{x^i}{i!}} \frac 1t\ dt$$

look at all feasible to me. Is there some theorem(s) that'd help make this a bit easier? Hints are welcome. :)

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    $\begingroup$ Perhaps you could try differentiating one of them and showing that the derivative is $1$? That would show that it's of the form $x+C$. $\endgroup$ – Akiva Weinberger Dec 14 '15 at 4:17
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Here's a sketch of the proof:

1) $f(x) = \exp x$ is differentiable everywhere with $f'(x) = f(x)$.

2) $f$ is invertible.

3) The inverse $f^{-1}$ has $(f^{-1})'(f) = 1/f$.

4) $f^{-1}(x) = \log x$.

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  • $\begingroup$ That last step took me a minute, but wow -- that's a nice way to do it. Thanks! $\endgroup$ – user298535 Dec 14 '15 at 4:28
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The power series for $f(x) = e^x$ makes it clear that $f'(x) = f(x)$ for all $x.$ From the FTC we see that if $g(x) = \ln x,$ then $g'(x) = 1/x.$ Consider the function $g(f(x)),$ defined for $x\in \mathbb R.$ The chain rule shows the derivative of this equals $g'(f(x))f'(x) = (1/f(x))f(x) = 1.$ Thus $g(f(x)) = x+c$ on $\mathbb R$ for some constant $c.$ Because $g(f(0)) = 0,$ we see $c=0.$

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