given { ((3^y −1= (6/2^x ))),(((3)^(y/x) = 2 )) :} find (1/x)+(1/y).


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Question Number 77186 by jagoll last updated on 04/Jan/20
$given { ((3^y −1= (6/2^x ))),(((3)^(y/x) = 2 )) :} find (1/x)+(1/y).$
$given$ ${\begin{cases} 3^{y} - 1 = \frac{6}{2^{x}} \\ {(3)}^{\frac{y}{x}} = 2 \end{cases} find \frac{1}{x} + \frac{1}{y} .$
	Answered by john santu last updated on 04/Jan/20
	$⇒ (3^y )= (2^x ) let 2^x = j ⇒ j^2 −j−6=0 (j−3)(j+2) =0 ⇒ { ((j=3 ⇒x=log_2 (3))),((3^(y ) = 3 ⇒ y = 1)) :} therefore (1/x)+(1/y) = log_3 (6)$
	$\Rightarrow (3^{y}) = (2^{x})$ $l e t 2^{x} = j \Rightarrow j^{2} - j - 6 = 0$ $(j - 3) (j + 2) = 0 \Rightarrow {\begin{cases} j = 3 \Rightarrow x = \log_{2} (3) \\ 3^{y} = 3 \Rightarrow y = 1 \end{cases}$ $t h e r e f o r e \frac{1}{x} + \frac{1}{y} = \log_{3} (6)$
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