Logarithmic function with strange bases

Question

Given $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n$.

I have rewritten $\log_{3n} 45$ as $\dfrac{\log_{4n}45}{\log_{4n}3n}$ and multiplied to get

$\log_{4n} 40\sqrt{3}\cdot\log_{4n}3n = \log_{4n} 45$

but do not know how to continue. Hints would be greatly appreciated, but please don't give me the answer. Thank you.

Answer 1

Hint:

$$\frac{\log40\sqrt{3}}{\log4n}=\log_{4n}40\sqrt{3}=\log_{3n}45=\frac{\log45}{\log3n}$$

Answer 2

Since $\log_ab=\frac{\ln b}{\ln a}$, where $\ln $ is the natural logarithm. So $\log_{4n} 40\sqrt{3} = \log_{3n} 45$ can be written as $$\log_{4n} 40\sqrt{3}=\frac{\ln 40\sqrt{3}}{\ln{4n} } = \log_{3n} 45=\frac{\ln 45}{\ln{3n} }$$. Then $$\frac{\ln 40\sqrt{3}}{\ln{n}+ \ln{4}}=\frac{\ln 45}{\ln{n}+ \ln{3} }.$$ Hence we can easily get $\ln n$ and then get n.