x-1-and-x-2-is-root-log-2-x-1-2-log-x-2-the-value-is-x-1-x-2-a-2-1-4-b-2-1-2-c-4-1-4-d-4-1-2-e-6-1-4-

Question Number 154989 by bagjagugum123 last updated on 24/Sep/21

x_{1} a n d x_{2} i s r o o t \log_{2} x^{(1 +^{2} \log x)} = 2, t h e v a l u e

i s x_{1} + x_{2} = \dots

a . 2 \frac{1}{4}

b . 2 \frac{1}{2}

c . 4 \frac{1}{4}

d . 4 \frac{1}{2}

e . 6 \frac{1}{4}

Answered by john_santu last updated on 24/Sep/21

\Leftrightarrow (\log_{2} x + 1) \log_{2} x = 2

l e t y = \log_{2} x

\Rightarrow y^{2} + y - 2 = 0

\Rightarrow (y + 2) (y - 1) = 0

{\begin{cases} y = - 2 \Rightarrow x_{1} = \frac{1}{4} \\ y = 1 \Rightarrow x_{2} = 2 \end{cases} \Rightarrow x_{1} + x_{2} = \frac{9}{4}

Commented by bagjagugum123 last updated on 24/Sep/21

t h a n k y o u v e r y m u c h S i r

Answered by puissant last updated on 24/Sep/21

\Rightarrow (1 + {l o g}_{2} x) {l o g}_{2} x = 2

t = {l o g}_{2} x \to t^{2} + t - 2 = 0

\Rightarrow t_{1} = - 2; t_{2} = 1

\frac{{l n x}_{1}}{l n 2} = - 2 \Rightarrow l n x = - 2 l n 2 \Rightarrow x = \frac{1}{4}

\frac{{l n x}_{2}}{l n 2} = 1 \Rightarrow {l n x}_{2} = l n 2 \Rightarrow x_{2} = 2

∴∵ x_{1} + x_{2} = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} . .

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