Prove that $\log(x^{\frac{n}{m}}) = \frac{n}{m}\log(x)$ I need to prove the following:
$$\log(x^{\frac{n}{m}}) = \frac{n}{m}\log(x)$$
In order to prove it I can only use the definition:
$$\log(x) = \int_1^x{\frac{1}{t}dt}$$
I tried using some change of variables, but I was not able to get to the result. 
Can anybody help me? 
 A: Given (defined):
$$\log(x^{m/n}) = \int_1^{x^{m/n}}{\frac{1}{t}dt}$$ where $n \ne 0$
Let $\displaystyle s = t^\frac{n}{m}$
$$\implies ds = \frac{n}{m}t^{\frac{n}{m}-1} dt$$
$$\implies ds = \frac{n}{m}\frac{s}{t} dt $$
$$\frac{m}{n}\frac{t}{s}ds = dt$$
Under the substitution and the new limits, we therefore get:
$$\Rightarrow \log(x^{m/n}) = \frac{m}{n}\int_1^x\frac{1}{t}\frac{t}{s}ds = \frac{m}{n}\int_1^x\frac{1}{s}ds = \frac{m}{n}\log(x)$$

Note that this can be extended to any real number $r \in \mathbb{R}$ and $r \ne 0$
Let $\displaystyle s = t^\frac{1}{r}$
$$\implies ds = \frac{1}{r}t^{\frac{1}{r}-1} dt$$
$$\implies ds = \frac{1}{r}\frac{s}{t} dt $$
$$r\frac{t}{s}ds = dt$$
Under the substitution and the new limits, we therefore get:
$$\Rightarrow \log(x^r) = r\int_1^x\frac{1}{t}\frac{t}{s}ds = r\int_1^x\frac{1}{s}ds = r\log(x)$$
A: If you know that
$$\log{(x^n)}=n\log{(x)}$$
For $n\in\mathbb{N}$ then we can say
$$m\log{(x^{\frac{n}m})}=\log{\left((x^{\frac{n}m})^m\right)}=\log{(x^n)}=n\log{(x)}$$
$$\therefore \log{(x^{\frac{n}m})}=\frac1m (m\log{(x^{\frac{n}m})})=\frac1m(n\log{(x)})=\frac{n}m\log{(x)}$$
