Rules of Logarithms 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. 
 A: 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.
A: 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. 
A: 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 $\ln(1)=0$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
