3^x 8^(x/(x+2)) =6


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Question Number 83759 by M±th+et£s last updated on 05/Mar/20

$3^{x} 8^{\frac{x}{x + 2}} = 6$
Commented by M±th+et£s last updated on 05/Mar/20

$s o l v e t h e e q u a t i o n$
Commented by niroj last updated on 06/Mar/20

$3^{x} . 8^{\frac{x}{x + 2}} = 6$ $3^{x} . 2^{3 (\frac{x}{x + 2})} = 6$ $3^{x} . 2^{\frac{3 x}{x + 2}} = 3.2$ $3^{x - 1} . 2^{\frac{3 x}{x + 2} - 1} = 1$ $3^{x - 1} . 2^{\frac{3 x - x - 2}{x + 2}} = {(3.2)}^{0}$ $3^{x - 1} . 2^{\frac{2 x - 2}{x + 2}} = {(3.2)}^{0}$ $3^{x - 1} . 2^{\frac{2 (x - 1)}{x - 1 + 3}} = 1$ $p u t x - 1 = a$ $3^{a} . 2^{\frac{2 a}{a + 3}} = 3^{0} . 2^{0}$ $3^{a} = 3^{0}, 2^{\frac{2 a}{a + 3}} = 2^{0}$ $a = 0, \frac{2 a}{a + 3} = 0$ $x - 1 = 0, 2 a = 0 \Rightarrow a = 0 \Rightarrow x - 1 = 0$ $x = 1 .$ $∴ t h e v a l u e o f x i s 1 .$
Commented by mr W last updated on 06/Mar/20

$i f a^{c} b^{d} = 1 t h e r e a r e t w o s o l u t i o n s :$ $s o l u t i o n 1 : c = d = 0$ $s o l u t i o n 2 : a^{c} = b^{- d} \Rightarrow c \ln a = - d \ln b \Rightarrow \frac{c}{d} = - \frac{\ln b}{\ln a}$
	Answered by behi83417@gmail.com last updated on 05/Mar/20
	$3^(x−1) =2^(1−((3x)/(x+2))) ⇒ { ((x−1=0⇒x=1)),((1−((3x)/(x+2))=0⇒^(x≠−2) 3x=x+2⇒x=1)) :}$
	$3^{x - 1} = 2^{1 - \frac{3 x}{x + 2}}$ $\Rightarrow {\begin{cases} x - 1 = 0 \Rightarrow x = 1 \\ 1 - \frac{3 x}{x + 2} = 0 \overset{x \neq - 2}{\Rightarrow} 3 x = x + 2 \Rightarrow x = 1 \end{cases}$
	Answered by mr W last updated on 05/Mar/20

	$3^{x - 1} 2^{\frac{2 (x - 1)}{x + 2}} = 1$ ${[3 \times 2^{\frac{2}{x + 2}}]}^{x - 1} = 1$ $x - 1 = 0 \Rightarrow x = 1$ $o r$ $3 \times 2^{\frac{2}{x + 2}} = 1$ $2^{\frac{2}{x + 1}} = \frac{1}{3}$ $\frac{2}{x + 2} = - \frac{\ln 3}{\ln 2}$ $\Rightarrow x = - (2 + \frac{2 \ln 2}{\ln 3}) \approx - 3.2619$
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