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I have come across the definition of a natural logarithm that says, $\ln x = \int_{1}^{x} \frac{1}{t} dt$.

Using this definition, I tried to prove a few rules. 'One' such rule is, $\ln x^r = r\ln x$, where $r$ is rational.( The nature of irrational numbers keeps us from the true value and this is why the rule does not apply to them, does it? )

Proof: $\frac{d}{dx} \ln x^r = \frac{r}{x} = r\frac{d}{dx} \ln x$

On taking the antiderivative we get, $\ln x^r= r\ln x + C$

For $x=1$, $C=0$.

This gives us $\ln x = \int_{1}^{x} \frac{1}{t} dt$. Now, $\frac{1}{t}$ is a continuous except at zero. This tells us that the integral function $\ln x$ is also continuous for $x>0$. The variable $x$ can take infinitely many values which are greater than zero. The rule above is true only for $x=1$. For $x=2$, we see that $C=\ln 2^r - \ln 2$. My calculator says that this difference is significant and that it cannot be ignored. Why are we not considering $C$ values for $x\neq 1$

Observation: I see that this procedure of making $C$ zero for a particular $x$ and then generalizing the rule is evident in the proofs of other rules.

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    $\begingroup$ This is not "the" definition of logarithm but rather "a" definition of the logarithm function. I admit it is the easiest and most convenient one for establishing properties of logarithm. Also the rule you have mentioned holds for all real numbers $r$ provided you have a suitable definition of $x^{r} $ when $x$ is irrational. One such definition is given by $x^{r} =\exp(r\log x) $ which is based on the rule you are trying to prove. $\endgroup$
    – Paramanand Singh
    Jul 23, 2017 at 8:24
  • $\begingroup$ And note that your proof does require $x=1$. You can't use any other value of $x$ to determine $C$. $\endgroup$
    – Paramanand Singh
    Jul 23, 2017 at 8:26
  • $\begingroup$ You will find a lot of interesting stuff about the natural logarithm here. $\endgroup$ Jul 23, 2017 at 8:47

4 Answers 4

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Actually you had a mistake for $x=2$, instead it should be:

$$C=\ln 2^r - \color{red} r\cdot \ln 2=0$$

Thus no matter what $x$ you choose, $C=0$.

EDIT

To add a comment for the question: notice that by getting the integration result, we already know that $C$ is a constant that once fixed, would work for all values of valid $x$. Thus we could find $C$ by letting it fit for $x=1$ or $x=2$, or other values. But once we find the $C$ by fitting the value of one $x$, it is guaranteed it fits for all $x$.

And for a particular value of $x$, we could just calculate it, instead of using the rule you want to prove here.

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  • $\begingroup$ Oh my! I realized that. But you see, we are trying to prove the rule here and while working out the constant, we are making use of the same rule. I find it a little confusing. $\endgroup$
    – R004
    Jul 23, 2017 at 3:27
  • $\begingroup$ @R004 So by the indefinite integration, we know that $C$ is a constant that is not changing with $x$ we take $\endgroup$
    – Jay Zha
    Jul 23, 2017 at 3:29
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    $\begingroup$ Now I realize. I made a very silly error with the constant. This helps. $\endgroup$
    – R004
    Jul 23, 2017 at 3:48
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I finally thought to post an answer here which is an extension of my comments.


We are given the definition $$\log x=\int_{1}^{x}\frac{dt}{t}\tag{1}$$ for all $x>0$. An immediate consequence of this definition is $$\frac{d} {dx}\log x=\frac{1}{x}\tag{2}$$ You are supposed to prove that $$\log x^{r} =r\log x\tag{3}$$ for rational $r$ and positive $x$. Your approach can be made a bit more rigorous by considering the function $g(x) =\log x^{r} - r\log x$ for all $x>0$. Note that $g(1)=0$ and $$g'(x) =\frac{rx^{r-1}}{x^{r}}-\frac{r}{x}=0$$ and hence by a simple corollary of mean value theorem the function $g$ is constant for all $x>0$ and thus $g(x)=g(1)=0$. This completes the proof of the rule.

Note that in order to get the constant value of the function $g$ there is no other option but to use the value $g(1)$. You can use the value $g(2)$ or $g(3)$ but using that it will never be possible to show that the constant value of $g$ is $0$.

Further the rule $(3)$ is valid for irrational values of $r$ also but it requires us to have a definition of the symbol $x^{r} $ for irrational $r$. One of the simplest approaches is to state that by definition the rule $(3)$ holds for irrational values of $r$ also. Thus we define $x^{r} =\exp(r\log x) $ for irrational $r$ where $\exp$ is the inverse of $\log $. In this approach the rule $(3)$ comes for free for all irrational values of $r$. You can choose any other definition of $x^{r} $ and based on that you can show that $(x^{r}) '=rx^{r-1}$ and then the proof of rule $(3)$ proceeds exactly as before.

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In the formula $\ln (x^r) =r\ln x +C$, the symbol $C$ is a constant. So regardless of which $x$ value you substitute you will get the same answer. We are simply picking the simplest one we can since $\ln(1)=0$.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \ln\pars{x^{r}} & \equiv \int_{1}^{x^{\large r}}{\dd t \over t} \,\,\,\stackrel{t\ =\ z^{\large r}}{=}\,\,\, \int_{1}^{x}{r\,z^{r - 1}\,\dd z \over z^{r}} = r\int_{1}^{x}{\dd z \over z} = r\ln\pars{x} \end{align}

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  • $\begingroup$ Perfect use of the substitution $t=z^{r} $. +1 $\endgroup$
    – Paramanand Singh
    Jul 24, 2017 at 7:52
  • $\begingroup$ @ParamanandSingh The straightforward way. $\endgroup$ Jul 24, 2017 at 19:46

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