<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] We could also take a "factorization" approach.  Since
$$ \log_{4n} 40\sqrt{3} \ \ =  \ \ \log_{3n} 45 \ \ = \ \ \alpha \ \ , $$
we can write
$$ (4n)^\alpha \ \ = \ \ 2^3·3^{1/2}·5 \ \ \ , \ \ \ (3n)^\alpha \ \ = \ \ 3^2·5 $$ $$  \Rightarrow \ \ n^\alpha \ \ = \ \ 2^{3 \ - \ 2·\alpha}·3^{1/2}·5 \ \ = \ \ (2^0)·3^{2 \ - \ \alpha}·5 \ \ .  $$
The "primes-raised-to-powers" in each of these factorizations need to "match", so we have
$$ 3 \ - \ 2·\alpha \ \ = \ \ 0 \ \ \ , \ \ \ \frac12 \ \ = \ \ 2 \ - \ \alpha \ \ , $$
both of which tell us that $ \ \alpha \ = \ \frac32 \ \ . $
Thus, we have $$ n^3 \ \ = \ \ (n^\alpha)^2 \ \ = \ \ (2^{3 \ - \ 2·\alpha}·3^{1/2}·5)^2 $$ $$ = \ \ 2^{6 \ - \ 4·\alpha}·3·25 \ \ = \ \ 2^{6 \ - \ 4·[3/2]}·3·25 \ \ = \ \ 2^0·75 $$
or
$$ n^3 \ \ = \ \   3^{4 \ - \ 2·\alpha}·5^2 \ \ = \ \ 3^{4 \ - \ 2·[3/2]}·25 \ \ = \ \ 3·25 \ \ . $$
[Since this problem has so few "moving parts", the posted answers end up being variations on the same concept.]
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