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## zmudz one year ago Given that $$\log_{4n} 40\sqrt{3} = \log_{3n} 45$$, find $$n^3$$.

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1. alekos

are you given an answer?

2. IrishBoy123
3. anonymous

zmudz u there?

4. phi

$\log_{4n} 40\sqrt{3} = \log_{3n} 45$ you could do this let $\log_{4n} 40\sqrt{3} = x =\log_{3n} 45$ so that we can say $\log_{4n} 40\sqrt{3} = x \\\log_{3n} 45 =x$ or in exponential form $(4n)^x = 40 \sqrt{3} \\ (3n)^x = 45$ divide the top equation by the bottom equation $\left(\frac{4}{3} \right)^x = \frac{2^3}{3^\frac{3}{2} }$ or, after squaring both sides $\left(\frac{4}{3} \right)^{2x} = \frac{4^3}{3^3} = \left( \frac{4}{3}\right)^3$ from which we find x= $$\frac{3}{2}$$ using $$(3n)^x = 45$$ we get $$3^3 n^3 = 45^2 = 5^2\cdot 3^4$$ and $n^3= \frac{5^2\cdot 3^4}{3^3} =75$

5. alekos

Truly Amazing! That's why you're on 99

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