x-log-2-4-x-2-x-1-4-x-

Question Number 170855 by mnjuly1970 last updated on 01/Jun/22

⌊ x ⌋ = {l o g}_{2} (4^{x} - 2^{x} - 1) \Rightarrow ⌊ 4^{x} ⌋ = ?

Answered by floor(10²Eta[1]) last updated on 01/Jun/22

⌊ x ⌋ = n \Rightarrow x = n + α, 0 ⩽ α < 1

\Rightarrow 2^{n} = 4^{n} . 4^{α} - 2^{α} 2^{n} - 1

4^{n} - 2^{n} - 1 ⩽ 2^{n} = 4^{n} . 4^{α} - 2^{α} 2^{n} - 1 < 4 . 4^{n} - 2 . 2^{n} - 1

2^{n} = y

y^{2} - 2 y - 1 ⩽ 0 \Rightarrow 1 - \sqrt{2} ⩽ y ⩽ 1 + \sqrt{2}

\Rightarrow y \in {1, 2}

{4 y}^{2} - 3 y - 1 > 0 \Rightarrow y < \frac{- 1}{4} e y > 1

\Rightarrow y > 1

\Rightarrow y = 2 \Rightarrow n = 1

\Rightarrow x = 1 + α, 0 ⩽ α < 1, ⌊ x ⌋ = 1

1 = \log_{2} (4^{1 + α} - 2^{1 + α} - 1)

\Rightarrow 3 = 4 . 4^{α} - 2 . 2^{α}, z = 2^{α}

{4 z}^{2} - 2 z - 3 = 0

1 ⩽ z = \frac{1 \pm \sqrt{13}}{4} = 2^{α} < 2

\Rightarrow 2^{α} = \frac{1 + \sqrt{13}}{4} \Rightarrow α = \log_{2} (\frac{1 + \sqrt{13}}{4})

4^{x} = 4^{1 + α} = 4 . 4^{\log_{2} (\frac{1 + \sqrt{13}}{4})} = 4 {(\frac{1 + \sqrt{13}}{4})}^{2}

\frac{7 + \sqrt{13}}{2} = 4^{x} \Rightarrow ⌊ 4^{x} ⌋ = 5

Commented by Tawa11 last updated on 02/Jun/22

Great sir

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