A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$. 
Any hints? Thanks!

My proposal was that
$$\frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ = \ \frac{40 \sqrt{3}}{45} \ = \ \frac{8}{3 \sqrt{3}} \ = \ (\frac{4}{3})^{3/2} \ . $$  
So both logarithms equal 3/2 and the rest follows as Zev Chonoles shows.
EDIT: Just to verify that the other piece checks,
$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)^3 \ \Rightarrow \ n^3 \ = \ \frac{4800}{64} \ = 75 \ . $$
