Question-137772

Question Number 137772 by peter frank last updated on 06/Apr/21

Answered by bemath last updated on 06/Apr/21

\Leftrightarrow \frac{{(2^{n})}^{3} + {(3^{n})}^{3}}{{(2^{n})}^{2} {(3)}^{n} + 2^{n} . {(3^{n})}^{2}} = \frac{7}{6}

l e t {\begin{cases} 2^{n} = u \\ 3^{n} = v \end{cases}

\Leftrightarrow \frac{u^{3} + v^{3}}{u^{2} v + {u v}^{2}} = \frac{7}{6}

\Leftrightarrow 6 u^{3} + 6 v^{3} = 7 u^{2} v + 7 {u v}^{2}

\Rightarrow u^{2} (6 u - 7 v) = v^{2} (7 u - 6 v)

\Rightarrow \frac{u^{2}}{v^{2}} = \frac{7 u - 6 v}{6 u - 7 u} = \frac{7 (\frac{u}{v}) - 6}{6 (\frac{u}{v}) - 7}

l e t \frac{u}{v} = ℓ \Rightarrow ℓ^{2} = \frac{7 ℓ - 6}{6 ℓ - 7}

\Leftrightarrow 6 ℓ^{3} - 7 ℓ^{2} - 7 ℓ + 6 = 0

(ℓ + 1) (6 ℓ^{2} - 13 ℓ + 6) = 0

ℓ = - 1 r e j e c t e d

(ℓ - \frac{3}{2}) (ℓ - \frac{2}{3}) = 0

n o w i t e a s y t o s o l v e

n = 1 o r n = - 1

Answered by Rasheed.Sindhi last updated on 07/Apr/21

\frac{{(2^{3})}^{n} + {(3^{3})}^{n}}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}

\frac{{(2^{n})}^{3} + {(3^{n})}^{3}}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}

\frac{(2^{n} + 3^{n}) (2^{2 n} - 2^{n} 3^{n} + 3^{2 n})}{6^{n} (2^{n} + 3^{n})} = \frac{7}{6}

6 . 2^{2 n} - 6 . 2^{n} 3^{n} + 6 . 3^{2 n} = 7 . 6^{n}

6 . 2^{2 n} - 13 . 2^{n} . 3^{n} + 6 . 3^{2 n} = 0

L e t 2^{n} = x & 3^{n} = y

6 x^{2} - 13 x y + 6 y^{2}

(3 x - 2 y) (2 x - 3 y) = 0

3 x = 2 y ∣ 2 x = 3 y

\frac{x}{y} = \frac{2}{3} ∣ \frac{x}{y} = \frac{3}{2}

\frac{2^{n}}{3^{n}} = {(\frac{2}{3})}^{1} ∣ \frac{2^{n}}{3^{n}} = {(\frac{2}{3})}^{- 1}

{(\frac{2}{3})}^{n} = {(\frac{2}{3})}^{1} ∣ {(\frac{2}{3})}^{n} = \frac{3}{2} = {(\frac{2}{3})}^{- 1}

n = 1 ∣ n = - 1