Deriving properties of logarithms from the definition $\ln(x) = \int_{1}^{x} \frac{\mathrm{d}t}{t}$ 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 $r$?
 A: $$\ln x^r = \int_1^{x^r}\frac{dt}{t}.$$
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.
A: $\ln(x^{0})=\ln 1=\displaystyle\int_{1}^{1}\dfrac{dt}{t}=0$.
$\ln(x^{-1})=\displaystyle\int_{1}^{x^{-1}}\dfrac{dt}{t}=-\int_{1}^{x}\dfrac{du}{u}=-\ln x$ 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 $x\mapsto\displaystyle\int_{1}^{x}\dfrac{dt}{t}$.
