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

I can't seem to find any identities to help me in this problem. Any hints or answers? Thanks in advance!

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set $\log_{4n} 40\sqrt{3} = \log_{3n} 45=t$, thus \begin{cases} 4^t n^t=40\sqrt{3}\\ 3^t n^t=45 \end{cases} we have $$\left(\frac{4}{3}\right)^t=\frac{8\sqrt{3}}{9}=\frac{8}{3\sqrt 3}\implies t=\frac 32$$ finally $$8n^{\frac 32}=40\sqrt 3$$ then $$n^3=75$$

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$$\dfrac{\log5+\dfrac{\log3}2+3\log2}{2\log2+\log n}=\dfrac{\log5+2\log3}{\log3+\log n}$$

On simplification,

$$3(2\log2-\log3)\log n=(2\log2-\log3)(2\log5-\log3)$$

$$\iff\log(n^3)=\log\dfrac{5^2}3$$ as $2\log2-\log3\ne0$

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By the given equality there exists $x$ such that $$(4n)^x=40\sqrt3\\(3n)^x=45$$ It follows $$(\frac 43)^x=\frac{8\sqrt3}{9}=(\frac{2}{\sqrt3)})^3\iff(\frac{2}{\sqrt3})^{2x}=(\frac{2}{\sqrt3)})^3\iff x=\frac32$$ Hence

$$(3n)^{\frac 32}=45\iff (3n)^3=(45)^2\iff n^3=\frac{(45)^2}{3^3}=75$$

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