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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.2012-12-13
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    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.2012-12-13
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    @Marvis: Can you be more specific? It doesn't seem like a trivial matter to me at all.2012-12-13
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    @aiao: Do you have a precise definition of $e$, or of what it means to take $e$ to a real number power?2012-12-13
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    @jonas _euler number_ it is a famous constant like $\pi$2012-12-13
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    @Ryan If $$f(x) = \int_1^x \dfrac{dt}t$$ and $$g'(x) = g(x)$$ with $g(0) = 1$. Setting $t = g(y)$, we get that $dt = g'(y) dy$. $$f(x) = \int_{0}^{g^{-1}(x)} \dfrac{g'(y) dy}{g(y)} = \int_{0}^{g^{-1}(x)} dy = g^{-1}(x)$$2012-12-13
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    @aiao: That is not an answer to my question. Analogously, $\log$ is a famous function, but as you can see from your question, it is important to consider precisely how it is defined if you are asking how to prove that it has certain properties. You have not said what $e$ means precisely, and perhaps more importantly, what "to the power $x$" means. E.g., what is $e^e$? (And I don't mean, "What are a few figures from its calculator decimal approximation?".)2012-12-13

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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 it2012-12-13
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    @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 $g(0) = 1$.2012-12-13
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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 $\frac{d}{dx} \ln(x) =\frac{1}{x}\log_e(e) = \frac{1}{x}$. It is because of this that $e$ holds special importance to calculus.

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.

Proving that $\frac{d}{dx}e^x =e^x$ is as easy as letting $y=e^x$ and evaluating $\frac{d}{dx}\ln(y) = \frac{d}{dx}x$ using the chain rule. It falls out kind of trivially.

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    I don't understand your objection to defining the natural log in this way. Once definited in the manner suggested by the OP, as far as I understand everything thing else could be retrieved from an albeit different starting point. I'm not saying it's better I just don't understand what's wrong with it that's all.2018-03-28
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    I think you misunderstand my objection. That is true you can start out differently and arrive at the same. But if it is true then why do it differently than how history has revealed it? What is the point in this arbitrary change? History has already established a particular development and doing it any other way disrespects that and neglects the historical education of mathematics. There may not be a practical difference in effect from a purely mathematical perspective, but in terms of the history and understanding the purpose motivating development, it is hugely impactful.2018-03-28
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    My objection isnt strictly with a change in definition - although that is partly true for reasons previously explained. Rather, my objection is with this common notion "mathematicians" seem to have that it makes sense to have multiple simultaneous "definitions" (which are man-made arbitrations) for a single notion, word, or concept. That doesnt work mathematically rigorous. You cannot guarantee two definitions are consistent with one another. One must be proven equivalent to another statement. Then its a theorem equivalence, not a defined one. For rigor you can only have one definition.2018-03-28
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    When and if you understand that fact, the question is begged which definition do you have/use? I see no practical reason to switch from the historical and traditional one, wherein students learn the history and conceptual development motivating the need, as they learn the concepts. When you switch definitions, they often become overly complicated and require advanced mathematics to prove theorems what ought to be much simpler for a lower-level student to understand from a more traditional approach. History has developed for us in increasing complexity. Changing that forces over-complication.2018-03-28