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Question Number 74168 by mr W last updated on 19/Nov/19

Commented by mr W last updated on 19/Nov/19

$s o l v e f o r x .$
	Answered by Rio Michael last updated on 19/Nov/19

	$d i v i d i n g b y 4^{x},$ $1 + {(\frac{6}{4})}^{x} = {(\frac{9}{4})}^{x}$ $1 + {(\frac{3}{2})}^{x} = {(\frac{3}{2})}^{2 x}$ $l e t {(\frac{3}{2})}^{x} = a$ $\Rightarrow 1 + a = a^{2}$ $a^{2} - a - 1 = 0$ $a = \frac{1 \pm \sqrt{5}}{2}$ $o r {(\frac{3}{2})}^{x} = \frac{1 \pm \sqrt{5}}{2}$ $x (l o g \frac{3}{2}) = l o g [\frac{1 \pm \sqrt{5}}{2}]$ $x = \frac{l o g [\frac{1 \pm \sqrt{5}}{2}]}{l o g \frac{3}{2}}$ $x \approx 1.2$
	Answered by Maclaurin Stickker last updated on 19/Nov/19
	$I just saw this question a few years ago in a pdf. 2^(2x) +2^x .3^x −3^(2x) =0 divide by 3^(2x) ((2/3))^(2x) +((2/3))^x −1=0 ((2/3))^x =y y^2 +y−1=0 y=((−1∓(√5))/2)⇒y_1 =((−1+(√5))/2) and y_2 =((−1−(√5))/2) ((2/3))^x =((−1−(√5))/2)⇒x∉R ((2/3))^x =((−1+(√5))/2)⇒x=log_(2/3) ((−1+(√5))/2) S={log_(2/3) (((−1+(√5))/2))}$
	$I j u s t s a w t h i s q u e s t i o n a f e w y e a r s$ $a g o i n a p d f .$ $2^{2 x} + 2^{x} . 3^{x} - 3^{2 x} = 0$ $d i v i d e b y 3^{2 x}$ ${(\frac{2}{3})}^{2 x} + {(\frac{2}{3})}^{x} - 1 = 0$ ${(\frac{2}{3})}^{x} = y$ $y^{2} + y - 1 = 0$ $y = \frac{- 1 \mp \sqrt{5}}{2} \Rightarrow y_{1} = \frac{- 1 + \sqrt{5}}{2} a n d y_{2} = \frac{- 1 - \sqrt{5}}{2}$ ${(\frac{2}{3})}^{x} = \frac{- 1 - \sqrt{5}}{2} \Rightarrow x \notin R$ ${(\frac{2}{3})}^{x} = \frac{- 1 + \sqrt{5}}{2} \Rightarrow x = {l o g}_{\frac{2}{3}} \frac{- 1 + \sqrt{5}}{2}$ $S = {{l o g}_{\frac{2}{3}} (\frac{- 1 + \sqrt{5}}{2})}$
	Commented by Maclaurin Stickker last updated on 19/Nov/19

	$I s t h e r e a n o t h e r s o l u t i o n t o t h e p r o b l e m ?$
	Commented by malwaan last updated on 20/Nov/19

	$y e s$ $c o m p l e x s o l u t i o n$
	Answered by MJS last updated on 20/Nov/19

	$4^{x} + 6^{x} - 9^{x} = 0$ $t = {(\frac{3}{2})}^{x} \land t > 0 \Leftrightarrow x = \frac{\ln t}{\ln \frac{3}{2}} = \frac{\ln t}{\ln 3 - \ln 2}$ $(- t^{2} + t + 1) t^{\frac{2 \ln 2}{\ln 3 - \ln 2}} = 0 \land t > 0$ $\Rightarrow t = \frac{1 + \sqrt{5}}{2}$ $\Rightarrow x = \frac{\ln (1 + \sqrt{5}) - \ln 2}{\ln 3 - \ln 2}$ $but if we allow x \in C$ $\Rightarrow t \in C$ $\Rightarrow$ $t_{1} = \frac{1 - \sqrt{5}}{2} \Rightarrow x_{1} = - x_{3} + \frac{π}{\ln 3 - \ln 2} i$ $t_{2} = 0 \Rightarrow x_{2} = - \infty$ $t_{3} = \frac{1 + \sqrt{5}}{2} \Rightarrow x_{3} = \frac{\ln (1 + \sqrt{5}) - \ln 2}{\ln 3 - \ln 2}$
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