I've been asked to show, using $\log_b{x} = \frac{\log_a{x}}{\log_a{b}}$ and that $\log_2{n} = O(n^\frac{1}{p})$ and $n^p = O(2^n)$, that $\log_b{n} = O(n^p)$.
I have no idea where to start for this, and I'm unsure of what method to use to prove it for all $b>1$ and $p>0$.
How do I prove this using what I've been told?
You have $\log_b n=O(\log_2 n)$, by the change of base formula for logarithm, and as $\log_2 n = O\bigl(n^\frac{1}{p}\bigr)$, there results by the usual rules of asymptotic analysis, that $$\log_b n = O\bigl(n^\frac{1}{p}\bigr)\qquad\text{for every }b>1.$$
Note: as we actually have $\log_b n=o(\log_2 n)$, there results a stronger result: $$\log_b n = o\bigl(n^\frac{1}{p}\bigr)\qquad\text{for every }b>1$$