<s>[INST] $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$. $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$.
I can't seem to find any identities to help me in this problem.  Any hints or answers?  Thanks in advance!
 [/INST] set $\log_{4n} 40\sqrt{3} = \log_{3n} 45=t$, thus
\begin{cases}
4^t n^t=40\sqrt{3}\\
3^t n^t=45
\end{cases}
we have
$$\left(\frac{4}{3}\right)^t=\frac{8\sqrt{3}}{9}=\frac{8}{3\sqrt 3}\implies t=\frac 32$$
finally
$$8n^{\frac 32}=40\sqrt 3$$
then
$$n^3=75$$
</s>