zmudz
  • zmudz
Given that \(\log_{4n} 40\sqrt{3} = \log_{3n} 45\), find \(n^3\).
Mathematics
chestercat
  • chestercat
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alekos
  • alekos
are you given an answer?
anonymous
  • anonymous
zmudz u there?

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phi
  • 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 \[ n^3= \frac{5^2\cdot 3^4}{3^3} =75\]
alekos
  • alekos
Truly Amazing! That's why you're on 99

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