A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$. 
Any hints? Thanks!
 A: Here's a kind of ugly way of doing it:
$$\begin{align*}
(12n)^{\log_{4n}(40\sqrt{3})}&=(12n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}(4n)^{\log_{4n}(40\sqrt{3})}&=(4)^{\log_{3n}(45)}(3n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}\cdot 40\sqrt{3}&=(4)^{\log_{3n}(45)}\cdot 45\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(4)^{\log_{3n}(45)-\frac{3}{2}}\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(3)^{\log_3(4)\cdot(\log_{3n}(45)-\frac{3}{2})}\\\\
\log_{4n}(40\sqrt{3})-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\log_{3n}(45)-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\end{align*}$$
Therefore
$$\begin{align*}
\log_{3n}(45)&=\frac{3}{2}\\\\
2025&=(3n)^3\\\\
n^3&=\frac{2025}{27}=75
\end{align*}$$
A: 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,
$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)^3 \ \Rightarrow \ n^3 \ = \ \frac{4800}{64} \ = 75 \ . $$
A: By elementary arithmetic operations (after / describing next action):
$$\log_{4n}40\sqrt{3}=\log_{3n}45\ \ \ \mbox{ / definition of logarithm}$$
$$(4n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } 4=\frac{4}{3}\cdot 3$$
$$\left({4\over 3}\cdot 3n\right)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / }(ab)^c=a^cb^c$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot (3n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } a^{\log_ab}=b$$
$$\left({4\over 3}\right)^{\log_{3n}45}\cdot 45=40\sqrt{3}\ \ \ \mbox { / }\cdot\frac{1}{45}$$
$$\left({4\over 3}\right)^{\log_{3n}45}={8\over 9}\sqrt{3}\ \ \ \mbox { / }\left({a\over b}\right)^c=\frac{a^c}{b^c}$$
$$\left({4\over 3}\right)^{\log_{3n}45}=\left({4\over 3}\right)^{3\over 2}\ \ \ \mbox { / }a^b=a^c\Rightarrow b=c\ \ (a\neq 0,1)$$
$$\log_{3n}45=\frac{3}{2}\ \ \ \mbox { / definition of logarithm}$$
$$(3n)^{3\over 2}=45\ \ \ \mbox { / powered by } \frac{2}{3}\mbox{ and divided by }3$$
$$n=\frac{\sqrt[3]{45^2}}{3}\ \ \ \mbox { / powered by 3}$$
$$n^3=\frac{45^2}{27}$$
$$n^3=75$$
