Question

Given that \(\log_{4n} 40\sqrt{3} = \log_{3n} 45\), find \(n^3\).

Answer 1

are you given an answer?

Answer 2

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Answer 3

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Answer 4

\[ \log_{4n} 40\sqrt{3} = \log_{3n} 45 \]
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
\[ n^3= \frac{5^2\cdot 3^4}{3^3} =75\]

Answer 5

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