Sum of the roots of the equation $9^{\log_{3}}^{(\log_{2} x)} = \log_{2} x - {(\log_{2} x)}^{2} + 1$ is:

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$2$

$\sqrt{2}$

$2 + 1 / \sqrt{2}$

$1 / 2$

answer is D.

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Detailed Solution

$9^{\log_{3}}^{(\log_{2} x)} = \log_{2} x - {(\log_{2} x)}^{2} + 1$

$\Rightarrow$ $3^{\log_{3} {(\log_{2} x)}^{2}} = \log_{2} x - {(\log_{2} x)}^{2} + 1$

$\Rightarrow$ ${(\log_{2} x)}^{2} = \log_{2} x - {(\log_{2} x)}^{2} + 1$ $(\log_{2} x > 0)$

$\Rightarrow$ $2 {(\log_{2} x)}^{2} - \log_{2} x - 1 = 0$

$\Rightarrow$ $\log_{2} x = - 1 / 2$ or $\log_{2} x = 1$

As $\log_{2} x > 0, \log_{2} x = 1$ or $x = 2$

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