<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] 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:
$$
\begin{align}&\log_{3n}{45^2}=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace .\end{align}
$$
</s>