Show that
using the integral defintion of the natural log we get
$\displaystyle \ln(x)=\int_{1}^{x}\frac{1}{t}dt$
let $\displaystyle u=t^{1/n} \iff u^n=t \to nu^{n-1}=dt$
$\displaystyle \ln(x^n)=\int_{1}^{x^n}\frac{1}{t}dt=\int_{1}^{x}\ frac{1}{u^n}(nu^{n-1}du)=n\int_{1}^{x}\frac{1}{u}du=n\ln(x)$
TheEmptySet you are a lifesaver,I have one further question,could you tell me how the limits of integration change from 1 and x^n to 1 and u? I would be very grateful(I kinda know its to do with the fact that you made a substitution)!