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Assuming the definition of the logarithmic function as follows: $$\ln{x}=\int_1^x \frac{dt}{t}$$ I managed to derive the formulas for logarithms as we know them: $$\ln(xy)=\int_1^{xy}\frac{dt}{t}=\int_1^x\frac{dt}{t}+\int_x^{xy}\frac{dt}{t}=\int_1^x\frac{dt}{t}+\int_1^y\frac{dt}{t}=\ln x+\ln y$$ By induction I could find, that $$\forall n\in \Bbb{N}:\ln{(x^n)}=n\ln{x}$$ I got stuck when I tried to find that $\ln{(x^n)}=n\ln{x}$ holds for all real $n$. My idea is following:

It is easy to check that $\ln'(x)=\frac{1}{x}$. By Leibniz integral rule, we find: $$\frac{d}{dx}\ln{x^r}=\frac{d}{dx}\int_1^{x^r}\frac{dt}{t}=\int_1^{x^r}\frac{\partial}{\partial x}\frac{dt}{t}+\frac{1}{x^r}\cdot rx^{r-1}=r\cdot\frac{1}{x}$$ So when I integrate back, I get: $$\int r\cdot\frac{1}{x}dx=r\int\frac{1}{x}dx=r\ln{x}$$ Can this be taken as a serious proof or I should be aiming to find something more rigorous? Could you please give me a hint on how to show that the property $\ln{(x^r)}=r\ln{x}$ holds for all reals $r$?

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    $\begingroup$ The main problem I have with your approach is that it's actually somewhat difficult to prove $\frac{d}{dx}x^r=rx^{r-1}$ for real $r$. The usual way is via logarithmic differentiation, which presupposes the natural log that you're trying to prove. $\endgroup$
    – Teepeemm
    Apr 12, 2018 at 18:09
  • $\begingroup$ Well, i can define $exp$ as an inverse of log and also it can be easily seen that $\ln'{x}=1/x$ also we need to know the chain rule which is then straightforward $\endgroup$ Apr 12, 2018 at 18:54

2 Answers 2

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$\ln(x^{0})=\ln 1=\displaystyle\int_{1}^{1}\dfrac{dt}{t}=0$.

$\ln(x^{-1})=\displaystyle\int_{1}^{x^{-1}}\dfrac{dt}{t}=-\int_{1}^{x}\dfrac{du}{u}=-\ln x$ by substitution that $u=1/t$.

$\ln(x^{-n})=\ln((x^{n})^{-1})=-\ln(x^{n})=-n\ln x$.

$\ln(x)=\ln((x^{1/q})^{q})=q\ln(x^{1/q})$, so $\ln(x^{1/q})=1/q\ln(x)$ for $q\in{\mathbb{N}}$.

$\ln(x^{p/q})=(p/q)\ln x$.

And finally we use the density of ${\mathbb{Q}}$ and the continuity of $x\mapsto\displaystyle\int_{1}^{x}\dfrac{dt}{t}$.

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  • $\begingroup$ Wow, such amazing approach, have never really came across this , so...If I prove a property for rationals, then because for every two rational numbers $p,q$ there exists some other $r$ in between then e.g. $\Bbb{Q}$ is dense and because my function is continuous, I can conclude that the property holds for all reals, is that correct? $\endgroup$ Apr 12, 2018 at 16:59
  • $\begingroup$ It should be that, for irrational $r$, find a sequence $(p_{n})$ of rational numbers such that $p_{n}\rightarrow r$, and we have $\ln(x^{r})=\lim_{n}\ln(x^{p_{n}})=\lim_{n}(p_{n}\ln(x))=r\ln(x)$. $\endgroup$
    – user284331
    Apr 12, 2018 at 17:01
  • $\begingroup$ We should note that this also relies on the continuity of $r \mapsto x^r$, the proof of which would depend on how we define $x^r$ for irrational $r$. One can define $x^r$ via a density argument along the lines of this answer, or one can show $\ln$ has a continuous inverse function $\exp: \mathbb R \to (0,\infty)$, define $x^r = \exp(r\ln x)$, and check that this definition has all the properties we want exponentiation to have. One benefit of the second approach is that this definition makes it easier to show $x^r$ is in fact differentiable in $r$. $\endgroup$
    – user360874
    Apr 12, 2018 at 19:00
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$$\ln x^r = \int_1^{x^r}\frac{dt}{t}.$$

Substitute $s^r = t$, so that $rs^{r-1}ds = dt$:

$$ \ln x^r = \int_1^{x}\frac{rs^{r-1}ds}{s^r} = r\int_1^{x}\frac{ds}{s} = r \ln x. $$

Edit

Of course, this relies on the property that $(x^r)' = rx^{r-1}$. To avoid circular reasoning, we have to derive this without using logarithms. For positive integers, it follows directly from the binomial expansion that

$$ (x^n)' = \lim_{h\rightarrow 0}\frac{(x+h)^n - x^n}{h} = nx^{n-1}. $$ For rational exponents, we can write $$ \left((x^{p/q})^q\right)' = (x^p)'. $$ Using the chain rule, it follows that $$ q(x^{p/q})^{q-1}(x^{p/q})' = px^{p-1}, $$ so that $$ (x^{p/q})' = (p/q)x^{p/q-1}, $$ and using continuity, we can extend this to $(x^r)' = rx^{r-1}$ for real exponents.

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  • $\begingroup$ I think we can't use that, in the derivation for differentiation of $x^r$ we use something we are trying to prove $\endgroup$ Apr 13, 2018 at 13:34
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    $\begingroup$ @MichalDvořák See my edit. $\endgroup$
    – Pulsar
    Apr 14, 2018 at 2:42

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