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You need a logarithm function to solve all power functions. That's a fact.

Power functions look like this: $f\colon x \mapsto a x^r \qquad a,r \in \mathbb{R}$

But why would you need a logarithm function to be able to solve such a function? Exponential functions do have an unknown exponent but they look different: $x \mapsto a^x$

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    I don't understand your question. What exactly do you mean by "solving all power functions", and in what sense do you "need" a logarithm function?2011-06-14
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    By "solve all power functions", do you mean "to **define** all power functions"? It is true that in order to be able to give a *general* definition that encompasses all possible real exponents, one usually uses logarithms. But it is possible to define rational exponents *without* relying on logarithms, and one can define arbitrary exponents as limits of rational exponents, so it is not quite "a fact" that you "need logarithms" to define these functions. If not what you meant, then what does it mean to "solve a function"?2011-06-14
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    And how is your question related to integration, which is what your title mentions? I confess to being confused.2011-06-14
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    I think the OP is asking why $\int x^{-1}\,dx$ involves a logarithm.2011-06-14
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    I have to admit I also don't really understand the question. Here is what I can say: an antiderivative of $x^\alpha$ is $\frac{x^{\alpha+1}-1}{\alpha+1}$. Now let $\alpha \to -1$ and watch how the miracle appears...2011-06-14
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    @Myself: Wow. I never would have guessed. Nowhere in the body do I see anything indicating that...2011-06-14
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    Why nobody is editing this to make it a clean meaningfull question?2011-06-15
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    @Arjang: I'm not doing it because I don't have a clear idea of what the OP is trying to ask. I could completely change the intended meaning, which is why I asked for clarification. Absent that clarification, I will not attempt a blind guess at the meaning.2011-06-15
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    @Arturo, cool dude, tx.2011-06-15

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The problem is this: you know that for $n \neq -1$, the indefinite integral of $x^n$ is equal to $\frac{x^{n+1}}{n+1}$. But this formula can't possibly be valid for $n = -1$ because the denominator vanishes. So instead you have to take the limit. It's easiest to see how this works with the definite integral

$$\int_a^b x^n \, dx = \frac{a^{n+1} - b^{n+1}}{n+1}.$$

If you want to see what happens at $n = -1$, what you do is to take the limit as $n \to -1$. By l'Hopital's rule, remembering that $x^k = e^{k \ln x}$, we find that

$$\lim_{n \to -1} \frac{a^{n+1} - b^{n+1}}{n+1} = \lim_{n \to -1} \frac{a^{n+1} \ln a - b^{n+1} \ln b}{1} = \ln a - \ln b.$$

So that's where the logarithm appears: it naturally comes out of the value of this limit, and in fact this limit can be used to define the logarithm.

More generally speaking, if you have a collection of functions closed under differentiation, you are in no way guaranteed that that collection of functions is also closed under integration. In fact given a class of functions, integration generally gives you new functions not in that class.

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    There is some circularity in using $\lim_{n\to-1}(a^{n+1}-1)/(n+1)=\log a$ to define the logarithm, when you're defining $a^{n+1}$ to be $e^{(n+1)\log a}$.2011-06-14
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    @Qiaochu , @Gerry, to simplify one can take a=1, then the limit will define the $\log (b)$2011-06-15
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    @Arjang, I did that (well, I took $b=1$), but my question remains: how can you use the limit to define the logarithm, when you've already used logarithms to define $a^{n+1}$?2011-06-15
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    @Gerry: if you use the above limit to define the logarithm, then naturally you aren't going to apply l'Hopital's rule. I just wanted to give a reason why the limit turns out the way it does that would make sense to someone with a standard calculus background.2011-06-15
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    @Qiaochu, l'Hopital's Rule isn't the problem, the problem is you defined $a^{n+1}$ using logarithms, and now you propose to define logarithms using $a^{n+1}$. Isn't that circular?2011-06-15
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    @Gerry: those are _separate_ explanations. All I'm trying to say is that on the one hand if you know some basic facts about exponentials and logarithms, you can deduce this result in a natural way; on the other hand, you can use this integral to define and prove basic facts about exponentials and logarithms.2011-06-15
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    @Qiaochu, if you are saying you can use $\int_1^x(1/t)\,dt$ to define and prove basic facts about logarithms, we have no disagreement. But in your answer you write "this limit" (which must refer to $\lim_{n\to-1}(a^{n+1}-b^{n+1})/(n+1)$) can be used to define the logarithm. "this limit" presupposes understanding of $a^{n+1}$ which is to say it presupposes understanding basic facts about exponentials; if you understand $a^{n+1}$ via $x^k=e^{k\log x}$ then it presupposes understanding logarithms. That, I maintain, is circular.2011-06-16
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    @Gerry: I hope we can agree that there is more than one way to define exponentials (for example defining them for integer exponents, then rational exponents, then extending by continuity). The remark about logarithms was, _as I have been saying_, a way to make the limit maximally sensible to evaluate for someone with a standard calculus background.2011-06-16