<s>[INST] Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:

Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).

My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) =  \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
 [/INST] $\displaystyle\frac{\ln(40\sqrt{3})}{\ln(4n)} = \frac{\ln(45)}{\ln(3n)}$
$\displaystyle \frac{\ln(3n)}{\ln(4n)}
= \frac{\ln(45)}{\ln(40\sqrt{3})} × \frac{k}{k}
= \frac{\ln(45^k)}{\ln[(40\sqrt{3})^k]}$
Find k, such that ratio inside ln/ln matched
$\displaystyle \frac{3n}{4n} = \left(\frac{45}{40\sqrt{3}}\right)^k = \left(\frac{3}{4}\right)^{3k/2}$
$→ k = \frac{2}{3}$
$3\,n = 45^{2/3}$
$27\,n^3 = 45^2$
$n^3 = 75$
</s>