$$f(x)=\log x\Leftrightarrow f^{-1}(x)=e^x.$$
Ok, $\log x$ is defined as the function $f(\cdot)$ such that: $f'(x)=\dfrac{1}{x}$. How to get, from this, the inverse of it $f^{-1}(x)$? And why $e=2.73\cdots$?
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actually just saying that $f'(x) = 1/x$ doesnt necassirly mean that $f = \ln(x)$. for all constants $c$ you have $(\ln(x) + c)' = \frac{1}{x}$ – mm-aops Jun 20 '14 at 22:10
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Another recent thread. – Raymond Manzoni Jun 22 '14 at 15:32
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1The approximation 2.73 for e is quite bad. At least use 2.72. – KCd Jul 04 '14 at 00:38
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The other answer has not answered your question, “Why is $e$ equal to $2.718281828\cdots$?”.
Let’s form the number $E=\lim_n(1+\frac1n)^n$, and evaluate it knowing the continuity of the log function and what its derivative is. Of course this number $E$ is computable, even if slowly, directly by hand. And if you take $n$ large enough, you will indeed find that your result is close to the number I quoted above.
We have: $$ \log(E)=\log\left[\lim_n(1+\frac1n)^n\right]=\lim_n\frac{\log(1+\frac1n)-\log(1)}{1/n}=\log'(1)=1/1\,, $$ just using the definition of the derivative and applying it to the logarithm. So $\log(E)=1$, and I think that’s what you wanted to know.
Lubin
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You also need to impose $\log(1) = 0$; otherwise, it's only determined up to a constant. The inverse $f^{-1}$ has derivative $$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = f^{-1}(x)$$ and $f^{-1}(0) = 1$ ; this is exactly the definition (well, one definition) of $e^x$.
anomaly
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