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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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