2^x − 3^x = (√(6^x − 9^x )) find: x = ?


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Question Number 200330 by hardmath last updated on 17/Nov/23

$2^{x} - 3^{x} = \sqrt{6^{x} - 9^{x}}$ $find : x = ?$
	Answered by Rasheed.Sindhi last updated on 17/Nov/23

	${(2^{x} - 3^{x})}^{2} = 6^{x} - 9^{x}$ $2^{2 x} - 2 (2^{x}) (3^{x}) + 3^{2 x} = 6^{x} - 9^{x}$ $4^{x} - 2 (6^{x}) + 9^{x} = 6^{x} - 9^{x}$ $4^{x} + 2 (9^{x}) = 3 (6^{x})$ ${(\frac{4}{6})}^{x} + 2 {(\frac{9}{6})}^{x} = 3$ ${(\frac{2}{3})}^{x} + 2 {(\frac{3}{2})}^{x} = 3$ $l e t {(\frac{2}{3})}^{x} = a$ $a + \frac{2}{a} = 3$ $a^{2} - 3 a + 2 = 0$ $(a - 1) (a - 2) = 0$ $a = 1, 2$ $∙ {(\frac{2}{3})}^{x} = 1 = {(\frac{2}{3})}^{0} \Rightarrow x = 0 ✓$ $∙ {(\frac{2}{3})}^{x} = 2$ $x \log_{2} (\frac{2}{3}) = \log_{2} 2 = 1$ $x = \frac{1}{\log_{2} 2 - \log_{2} 3} = \frac{1}{1 - \log_{2} 3} ✓$
	Commented by Rasheed.Sindhi last updated on 18/Nov/23

	$I a l w a y s r e m e m b e r a n o l d f o r u m$ $m e m b e r^{'} {d i n u s a r}^{'},^{'} {h o n g k i n g}^{'};$ $w h e n I s e e y o u r p o s t s$
	Commented by hardmath last updated on 18/Nov/23

	$thank you dear professor$
	Answered by Rasheed.Sindhi last updated on 17/Nov/23

	$2^{x} - 3^{x} = \sqrt{6^{x} - 9^{x}}$ $2^{x} - 3^{x} = 3^{\frac{x}{2}} \sqrt{2^{x} - 3^{x}}$ $2^{x} - 3^{x} = 3^{\frac{x}{2}} \sqrt{2^{x} - 3^{x}}$ ${(\sqrt{2^{x} - 3^{x}})}^{2} - 3^{\frac{x}{2}} \sqrt{2^{x} - 3^{x}}$ $\sqrt{2^{x} - 3^{x}} (\sqrt{2^{x} - 3^{x}} - 3^{x / 2}) = 0$ $\sqrt{2^{x} - 3^{x}} = 0 \lor \sqrt{2^{x} - 3^{x}} = 3^{x / 2}$ $▸ 2^{x} = 3^{x} \Rightarrow {(\frac{2}{3})}^{x} = 1 = {(\frac{2}{3})}^{0} \Rightarrow x = 0$ $▸ \sqrt{2^{x} - 3^{x}} = 3^{x / 2}$ $2^{x} - 3^{x} = 3^{x}$ $2^{x} = 2 \cdot 3^{x}$ ${(\frac{2}{3})}^{x} = 2$ $\log_{2} {(\frac{2}{3})}^{x} = \log_{2} 2$ ${xlog}_{2} (\frac{2}{3}) = 1$ $x = \frac{1}{\log_{2} 2 - \log_{2} 3} = \frac{1}{1 - \log_{2} 3}$
	Answered by witcher3 last updated on 17/Nov/23

	$\sqrt{1 (1 - {(\frac{3}{2})}^{x}} = {(\frac{2}{3})}^{\frac{x}{2}} - {(\frac{3}{2})}^{\frac{x}{2}}$ $\sqrt{1 - a^{2}} = (\frac{1}{a}) - a$ $a = 1; x => p 0; or a = \sqrt{1 - a^{2}}$ $\Leftrightarrow a = \frac{1}{\sqrt{2}} \Rightarrow {(\frac{3}{2})}^{\frac{x}{2}} = \frac{1}{\sqrt{2}}$ $\Leftrightarrow x = 2 \frac{\ln (\frac{1}{\sqrt{2}})}{\ln (\frac{3}{2})}$
	Answered by manxsol last updated on 18/Nov/23


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