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My math professor told me that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$ by the definition; so far so good.

But how/why does $\ln(x)$ ($\int_1^x\frac{1}{t} dt$: by defintion) coincide with the inverse of $e^{x}$? Thanks!

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You mean more specifically that $\ln(x)=\int_1^x\frac{1}{t}dt$. Well, is $e^x$ not the inverse of $\int_1^x\frac{1}{t}dt$ by definition? If not, please share what the definition of $e^x$ is. – Jonas Meyer Dec 13 '12 at 6:17
What is your definition of $e^x$? If you define $e^x$ as the function $f(x)$ such that $f'(x) = f(x)$ with $f(0) = 1$, then what you want follows immediately from change of variables. – user17762 Dec 13 '12 at 6:18
@Marvis: Can you be more specific? It doesn't seem like a trivial matter to me at all. – Ryan Dec 13 '12 at 6:26
@aiao: Do you have a precise definition of $e$, or of what it means to take $e$ to a real number power? – Jonas Meyer Dec 13 '12 at 6:26
@jonas euler number it is a famous constant like $\pi$ – aiao Dec 13 '12 at 6:29
up vote 4 down vote accepted

If $$f(x) = \int_1^x \dfrac{dt}t$$ and $$g'(x) = g(x)$$ with $g(0) = 1$, then we can show that $f(x) = g^{-1}(x)$.

Setting $t = g(y)$, we get that $dt = g'(y) dy$. When $y=0$, we get that $t=1$ and when $y = g^{-1}(x)$, we get that $t=x$.

Hence, $$f(x) = \int_{0}^{g^{-1}(x)} \dfrac{g'(y) dy}{g(y)} = \int_{0}^{g^{-1}(x)} dy = g^{-1}(x)$$

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are you sure that the limits in the last equation are rights? ($g(0)$?).... edit: my bad.... Got it – aiao Dec 13 '12 at 6:37
@aiao $g(0) = 1$. Note that $g'(x) = g(x)$ has infinite functions as solutions unless you specify the value of $g(0)$. $e^x$ is the function such that $g'(x) = g(x)$ and $g(0) = 1$. – user17762 Dec 13 '12 at 6:40

I disagree/disapprove of any math teacher that says this is definition. Its not. No one can assign more than one definition to a single concept without somewhere proving the equivalence.

You want to talk definition? One of the Bernoulli family, while working on problems of bank account growth and compounding interest problems, was able to show that a bank account that grows with 100% APR, compounded continuously, converged to a value a little over 2.7 times that of the principle. HE showed that $\lim_{n\to \infty} (1+\frac{1}{n})^n = e$.

THIS is definition. The most fundamental and principle of equalities. This is the historical origin, the chronological first, and the basis of all other properties which followed, and which were proved equivalent.

So now youve established the existence of e. Taking $e^x$ is a trivial algebraic concept. Taking example from the interest growth problems, $e^x$ is the balance of a bank account with a principle of 1 after $x$ years.

Enter logarithms: If $e^m = n $ then (by definition of the logarithm) we have $m = \log_e(n)$.

Then, we can say $\log_e(n) = \ln(n)$. This is nomenclature. Its just a standard and simpler way of writing a frequently appearing logarithm. This equivalence is not proven, its definition. But its not a mathematical coincidence, its simply an arbitrated truth. It was assigned and invented for simplicity.

Proving $\frac{d}{dx} \ln(x) = \frac{1}{x}$ is done via the limit definition of the derivative. Here, let us prove the derivative of any arbitrary logarithm with respect ot its argument x. $\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{\log_a(x+h)-\log_a(x)}{h}$

$\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{\log_a(\frac{x+h}{x})}{h}$

$\frac{d}{dx} \log_a(x) = \lim_{h\to 0}\frac{1}{h}\log_a(1+\frac{h}{x})$

Letting $h=x/n$ then as $h\to 0$ we have $n\to\infty$. The substitution creates:

$\frac{d}{dx} \log_a(x) = \lim_{n\to \infty}\frac{n}{x}\log_a(1+\frac{1}{n})$

Factoring out constants and continuous functions independent of $n$: $\frac{d}{dx} \log_a(x) = \frac{1}{x}\log_a\lim_{n\to \infty}(1+\frac{1}{n})^n$

By the ACTUAL definition of e, we have $\frac{d}{dx} \log_a(x) = \frac{1}{x}\log_a(e)$

This is a general truth for any log. If you let $a=e$ then $\frac{d}{dx} \ln(x) =\frac{1}{x}\log_e(e) = \frac{1}{x}$

And by the general theorem of calculus, we also have the integral, $\ln(x) = \int_1^x \frac{1}{t} dt$. This is just the reverse rule.

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