Changing the logarithm base on a linearised graph

The variables \(x\) and \(y\) satisfy \(y=ax^n\), where \(a\) and \(n\) are positive constants.

1. Show that, for any base \(p>0,\ p\ne1\), the graph of \(\log_p y\) against \(\log_p x\) is a straight line, and state its gradient and intercept.[2]
2. One student plots \(\ln y\) against \(\ln x\) and obtains gradient \(1.5\) and intercept \(1.386\). Find \(a\) and \(n\).[2]
3. Another student plots \(\lg y\) against \(\lg x\) for the same data. Predict her gradient and intercept, and explain why the gradient does not depend on the base but the intercept does.[3]