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?

Alternative approach:
Let $~r~$ denote $\displaystyle \log_{4n} 40\sqrt{3}.$
Then
$$(4n)^r = 40\sqrt{3}, ~~(3n)^r = 45 \implies $$
$$\left[\frac{4}{3}\right]^r = \left[\frac{4n}{3n}\right]^r = \frac{40\sqrt{3}}{45} = \frac{8\sqrt{3}}{9} \implies $$
$$9 \times 4^r = 8\sqrt{3} \times 3^r \implies $$
$$\frac{2^{2r}}{8} = \frac{\sqrt{3} \times 3^r}{9} \implies $$
$$2^{2r - 3} = 3^{r - (3/2)} = \sqrt{3}^{2r-3}. \tag1 $$
The only way that this is possible is if $(2r - 3) = 0 \implies r = (3/2).$
Checking this candidate value for $r$ gives
$$4^{3/2} \times n^{3/2} = 40\sqrt{3} \implies n^{3/2} = 5\sqrt{3}.$$
Checking this further,
$$3^{3/2} \times 5\sqrt{3}$$
does in fact equal $45$.
So, the sole candidate value for $r$, works.
Therefore,
$$n^{3/2} = 5\sqrt{3} \implies n^3 = \left[5\sqrt{3}\right]^2 = 75.$$
