Scan0001-1.jpg - Image - Photobucket - Video and Image Hosting
Any hints or anything to steer me in the right direction would be awesome!
Scan0001-1.jpg - Image - Photobucket - Video and Image Hosting
Any hints or anything to steer me in the right direction would be awesome!
Well, we know that $\displaystyle \sqrt[n]{a^m} = a^{\frac{m}{n}}$.
So we just have to prove that: $\displaystyle \log_{b} a^{\frac{m}{n}} = \frac{m}{n} \log_{b} a$
Let: $\displaystyle {\color{red}u = \log_{b}a} \ \iff \ b^u = a$
Raise both sides to the $\displaystyle m/n$: $\displaystyle \left(b^u\right)^{\displaystyle m/n} = a^{\displaystyle m/n} \ \iff \ b^{\displaystyle mu/n} = a^{\displaystyle m/n} \ \iff \ \log_{b} \left(a^{\displaystyle m/n}\right) = \frac{m{\color{red}u}}{n} = \hdots$