Hello everyone I try to solve 4^(x+1) +2^(2−x) =65 Thx in advance


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Question Number 194176 by Skabetix last updated on 29/Jun/23

$H e l l o e v e r y o n e$ $I t r y t o s o l v e 4^{x + 1} + 2^{2 - x} = 65$ $T h x i n a d v a n c e$
	Answered by Skabetix last updated on 29/Jun/23

	$I f o u n d x = 2 b u t t h e r e i s a n o t h e r s o l u t i o n$
	Answered by mr W last updated on 29/Jun/23

	$4 {(2^{x})}^{2} + \frac{4}{2^{x}} = 65$ $4 {(2^{x})}^{3} - 65 (2^{x}) + 4 = 0$ $l e t u = 2^{x}$ $4 u^{3} - 65 u + 4 = 0$ $4 u^{3} - 16 u^{2} + 16 u^{2} - 64 u - u + 4 = 0$ $(u - 4) (4 u^{2} + 16 u - 1) = 0$ $\Rightarrow u = 4 = 2^{x} \Rightarrow x = 2 ✓$ $\Rightarrow u = \frac{- 4 + \sqrt{17}}{2} = 2^{x} \Rightarrow x = \log_{2} \frac{- 4 + \sqrt{17}}{2} ✓$
	Answered by MM42 last updated on 29/Jun/23

	$2^{x} (2^{2 x + 2} + 2^{2 - x} = 65)$ $4 \times 2^{3 x} + 4 = 65 \times 2^{x}$ $2^{x} = y$ $4 y^{3} - 65 y + 4 = 0$ $(y - 4) (4 y^{2} + 16 y - 1) = 0$ $y = 4 = 2^{x} \Rightarrow x = 2$ $y = - 8 \pm \sqrt{68} = - 8 \pm 2 \sqrt{17} = 2 (- 4 \pm \sqrt{17})$ $2^{x} = 2 (- 4 \pm \sqrt{17}) \Rightarrow x = 2 + {l o g}_{2} (- 4 \pm \sqrt{17})$
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