The sum of the roots of the equation, x + 1 - 2 _ 2 (3 + 2 ^ x ) +...

Given, $x + 1 - 2\log_{2}\left(3 + {2}^{x}\right) + 2\log_{4}\left(10 - {2}^{- x}\right) = 0$

$⇒ x + 1 - 2\log_{2}\left(3 + {2}^{x}\right) + 2log{2}^{2}\left(\dfrac{10 . {2}^{x} - 1}{{2}^{x}}\right) = 0$

$⇒ x + 1 - 2\log_{2}\left(3 + {2}^{x}\right) + \log_{2}\left(\dfrac{10 . {2}^{x} - 1}{{2}^{x}}\right) = 0$

$⇒ x + 1 - 2\log_{2}\left(3 + {2}^{x}\right) + \log_{2}\left(10 . {2}^{x} - 1\right) - \log_{2}{2}^{x} = 0$

$⇒ x + 1 - \log_{2}{\left(3 + {2}^{x}\right)}^{2} + \log_{2}\left(10 . {2}^{x} - 1\right) - x{\log}_{2}2 = 0$

$⇒ x + 1 - \log_{2}{\left(3 + {2}^{x}\right)}^{2} + \log_{2}\left(10 . {2}^{x} - 1\right) - x = 0$

$⇒ x + 1 + \log_{2}\left[\dfrac{10 . {2}^{x} - 1}{{\left(3 + {2}^{x}\right)}^{2}}\right] - x = 0$

$⇒ 1 + \log_{2}\left[\dfrac{10 . {2}^{x} - 1}{{\left(3 + {2}^{x}\right)}^{2}}\right] = 0$

$⇒ \log_{2}\left[\dfrac{10 . {2}^{x} - 1}{{\left(3 + {2}^{x}\right)}^{2}}\right] = - 1$

$⇒ \log_{2}\left[\dfrac{10 . {2}^{x} - 1}{9 + {\left({2}^{x}\right)}^{2} + 6 . {2}^{x}}\right] = \log_{2}\left(\dfrac{1}{2}\right)$

$⇒ \dfrac{10 . {2}^{x} - 1}{9 + {\left({2}^{x}\right)}^{2} + 6 . {2}^{x}} = \dfrac{1}{2}$

$⇒ 2\left(10 . {2}^{x} - 1\right) = 9 + {\left({2}^{x}\right)}^{2} + 6 . {2}^{x}$

$⇒ {\left({2}^{x}\right)}^{2} - 14 . {2}^{x} + 11 = 0$

Let ${2}^{x} = y$

$⇒ {y}^{2} - 14y + 11 = 0$

Let roots are ${y}_{1} = {2}^{{x}_{1}} \text{ and } {y}_{2} = {2}^{{x}_{2}}$

Product of roots, ${y}_{1}{y}_{2} = {2}^{{x}_{1}} \times {2}^{{x}_{2}} = {2}^{{x}_{1} + {x}_{2}} = 11$

$⇒ log{2}^{{x}_{1} + {x}_{2}} = \log 11$

$⇒ \left({x}_{1} + {x}_{2}\right)\log 2 = \log 11$

$⇒ {x}_{1} + {x}_{2} = \dfrac{\log 11}{\log 2}$

Hence, ${x}_{1} + {x}_{2} = \log_{2}11$