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?
 A: I would go the following way.
You have :
$$\begin{align}&\log_{4n} 40\sqrt{3} = \log_{3n} 45\\
\implies &\log_{4n} \left(3\cdot 40^2\right) = \log_{3n} 45^2=k
\end{align}$$
This leads to :
$$\begin{align}&\begin{cases}(4n)^k=3\cdot 40^2\\ 
(3n)^k=45^2 \end{cases}\\
\implies &\left(\frac 43\right)^k=\frac {3\cdot 40^2}{45^2}=\left(\frac {4}{3}\right)^3\\
\implies &k=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace.\end{align}$$
A: $\displaystyle\frac{\ln(40\sqrt{3})}{\ln(4n)} = \frac{\ln(45)}{\ln(3n)}$
$\displaystyle \frac{\ln(3n)}{\ln(4n)}
= \frac{\ln(45)}{\ln(40\sqrt{3})} × \frac{k}{k}
= \frac{\ln(45^k)}{\ln[(40\sqrt{3})^k]}$
Find k, such that ratio inside ln/ln matched
$\displaystyle \frac{3n}{4n} = \left(\frac{45}{40\sqrt{3}}\right)^k = \left(\frac{3}{4}\right)^{3k/2}$
$→ k = \frac{2}{3}$
$3\,n = 45^{2/3}$
$27\,n^3 = 45^2$
$n^3 = 75$
A: 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,
$$n^{3/2} = 5\sqrt{3} \implies n^3 = \left[5\sqrt{3}\right]^2 = 75.$$
A: 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.]
A: In this answer, we provide another method that uses the definition and basic properties of logarithm.
The original equation states :
$$
\begin{align}&\log_{4n}{40}\sqrt{3}=\log_{3n}{45}\\
\implies &{40}\sqrt 3=\left(4n\right)^{\log_{3n}45}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=\left(3n\cdot \frac 43\right)^{\log_{3n}{45}}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=45\cdot \left(\frac 43\right)^{\log_{3n}{45}}\end{align}
$$
This leads to the following :
$$
\begin{align}\left(\frac {2}{\sqrt 3}\right)^{2\log_{3n}{45}}&=\frac {40\sqrt 3}{45}=\left(\frac {2}{\sqrt 3}\right)^3\end{align}
$$
Finally, again using the definition of logarithm, we get the desired result:
$$
\begin{align}&\log_{3n}{45^2}=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace .\end{align}
$$
