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Any hints or anything to steer me in the right direction would be awesome!

2. By law of indices

nth root of (x to the power of m) becomes x to the power of m/n

gives log_b x^(m/n)
Then by

Power Law of Logarithms:
log a^x = x log a

we get m/nlog_b x

3. I have to prove the law, I can not rely on it. Also where does the b^t play in?
We apparently have to use it.

4. I'm pretty much stuck with jaydee323 here.

5. Well, we know that $\sqrt[n]{a^m} = a^{\frac{m}{n}}$.

So we just have to prove that: $\log_{b} a^{\frac{m}{n}} = \frac{m}{n} \log_{b} a$

Let: ${\color{red}u = \log_{b}a} \ \iff \ b^u = a$

Raise both sides to the $m/n$: $\left(b^u\right)^{\displaystyle m/n} = a^{\displaystyle m/n} \ \iff \ b^{\displaystyle mu/n} = a^{\displaystyle m/n} \ \iff \ \log_{b} \left(a^{\displaystyle m/n}\right) = \frac{m{\color{red}u}}{n} = \hdots$