Deriving properties of logarithms from the definition $\ln(x) = \int_{1}^{x} \frac{\mathrm{d}t}{t}$

Question

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 $r$?

Answer 1

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

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

Answer 2

$$\ln x^r = \int_1^{x^r}\frac{dt}{t}.$$

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.