find-the-possible-values-of-x-if-8-x-27-x-12-x-18-x-7-6-

Question Number 19063 by chux last updated on 03/Aug/17

find the possible values of x if

\frac{8^{x} + 27^{x}}{12^{x} + 18^{x}} = \frac{7}{6}

Answered by 433 last updated on 03/Aug/17

\frac{2^{3 x} + 3^{3 x}}{{(3 \times 4)}^{x} + {(9 \times 2)}^{x}} = \frac{7}{6}

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

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

2^{x} = u & 3^{x} = v

\frac{u^{3} + v^{3}}{uv (u + v)} = \frac{7}{6}

{6 u}^{3} + {6 v}^{3} = {7 u}^{2} v + {7 uv}^{2}

6 {(u + v)}^{3} = 25 uv (u + v)

(u + v) (6 {(u + v)}^{2} - 25 uv) = 0

u + v = 0 \Leftrightarrow 2^{x} + 3^{x} = 0 *

6 {(u + v)}^{2} - 25 uv = 0

{6 u}^{2} + 12 uv + {6 v}^{2} - 25 uv = 0

{6 u}^{2} - 13 uv + {6 v}^{2} = 0

{6 u}^{2} - 4 uv - 9 uv + {6 v}^{2} = 0

2 u (3 u - 2 v) - 3 v (3 u - 2 v) = 0

(2 u - 3 v) (3 u - 2 v) = 0

2 u = 3 v \Leftrightarrow 2 \times 2^{x} = 3 \times 3^{x} \Leftrightarrow 2^{x + 1} = 3^{x + 1} \Leftrightarrow {(\frac{2}{3})}^{x + 1} = {(\frac{2}{3})}^{0} \Leftrightarrow x = - 1

3 u = 2 v \Leftrightarrow 3 \times 2^{x} = 2 \times 3^{x} \Leftrightarrow 2^{x - 1} = 3^{x - 1} \Leftrightarrow {(\frac{2}{3})}^{x - 1} = {(\frac{2}{3})}^{0} \Leftrightarrow x = 1

Commented by NEC last updated on 04/Aug/17

h m m m m

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

s i r p l s i d n t u n d e r s t a n h o w u g o t t h a t 6^{x}

Commented by 433 last updated on 04/Aug/17

{(3 \times 4)}^{x} + {(9 \times 2)}^{x} = 3^{x} 2^{x} 2^{x} + 3^{x} 3^{x} 2^{x} = 3^{x} 2^{x} (2^{x} + 3^{x}) = 6^{x} (2^{x} + 3^{x})

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

w a a w b r i l l i a n t s i r t h a n k z i c o m p l y n o w

Answered by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17

2^{x} = a, 3^{x} = b, a + b \neq 0 \Rightarrow \frac{a^{3} + b^{3}}{a^{2} b + {a b}^{2}} = \frac{7}{6} \Rightarrow

6 (a^{2} - a b + b^{2}) = 7 a b \Rightarrow 6 a^{2} - 13 a b + 6 b^{2} = 0

a = \frac{13 b \pm \sqrt{169 b^{2} - 144 b^{2}}}{12} = \frac{13 b \pm 5 b}{12} = (\frac{3}{2}, \frac{2}{3}) b

\Rightarrow {\begin{cases} \frac{a}{b} = \frac{2}{3} \Rightarrow {(\frac{2}{3})}^{x} = \frac{2}{3} \Rightarrow x = 1 \\ \frac{a}{b} = \frac{3}{2} \Rightarrow {(\frac{2}{3})}^{x} = \frac{3}{2} \Rightarrow x = - 1 . \end{cases}

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

h m m s i r a m b a f f l e a b o u t h o w u g o t t h a t 6 (a^{2} - a b + b^{2})

Commented by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17

\frac{a^{3} + b^{3}}{a^{2} b + {a b}^{2}} = \frac{(a + b) (a^{2} - a b + b^{2})}{a b (a + b)} = \frac{a^{2} - a b + b^{2}}{a b}

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

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