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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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    @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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    @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