The sum of the roots of the equation

$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$, is :

$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$<br><br>$${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$$<br><br>$$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$$<br><br>${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1$<br><br>$\Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}$<br><br>$\therefore$ ${({2^x})^2} - 14({2^x}) + 11 = 0$<br><br>Roots are 2^x₁ &amp; 2^x₂<br><br>$\therefore$ 2^x₁ . 2^x₂ = 11<br><br>x₁ + x₂ = log₂(11)