The sum of the roots of the equation, x+1-2 log_2(3+2^x)+2 log_4(10-2^-x)=0, is :

Given, $x+1-2{\mathrm{log}}_{2}(3+{2}^{x})+2{\mathrm{log}}_{4}(10-{2}^{-x})=0$x+1−2log2​(3+2x)+2log4​(10−2−x)=0
$\Rightarrow x+1-2{\mathrm{log}}_{2}(3+{2}^{x})+2{\mathrm{log}}_{{2}^{2}}(\frac{10.{2}^{x}-1}{{2}^{x}})=0$⇒x+1−2log2​(3+2x)+2log22​(2x10.2x−1​)=0
$\Rightarrow x+1-2{\mathrm{log}}_{2}(3+{2}^{x})+{\mathrm{log}}_{2}(\frac{10.{2}^{x}-1}{{2}^{x}})=0$⇒x+1−2log2​(3+2x)+log2​(2x10.2x−1​)=0
$\Rightarrow x+1-2{\mathrm{log}}_{2}(3+{2}^{x})+{\mathrm{log}}_{2}(10.{2}^{x}-1)-{\mathrm{log}}_{2}{2}^{x}=0$⇒x+1−2log2​(3+2x)+log2​(10.2x−1)−log2​2x=0
$\Rightarrow x+1-{\mathrm{log}}_{2}{(3+{2}^{x})}^{2}+{\mathrm{log}}_{2}(10.{2}^{x}-1)-x{\mathrm{log}}_{2}2=0$⇒x+1−log2​(3+2x)2+log2​(10.2x−1)−xlog2​2=0
$\Rightarrow x+1-{\mathrm{log}}_{2}{(3+{2}^{x})}^{2}+{\mathrm{log}}_{2}(10.{2}^{x}-1)-x=0$⇒x+1−log2​(3+2x)2+log2​(10.2x−1)−x=0
$\Rightarrow x+1+{\mathrm{log}}_{2}[\frac{10.{2}^{x}-1}{{(3+{2}^{x})}^{2}}]-x=0$⇒x+1+log2​[(3+2x)210.2x−1​]−x=0
$\Rightarrow 1+{\mathrm{log}}_{2}[\frac{10.{2}^{x}-1}{{(3+{2}^{x})}^{2}}]=0$⇒1+log2​[(3+2x)210.2x−1​]=0
$\Rightarrow {\mathrm{log}}_{2}[\frac{10.{2}^{x}-1}{{(3+{2}^{x})}^{2}}]=-1$⇒log2​[(3+2x)210.2x−1​]=−1
$\Rightarrow {\mathrm{log}}_{2}[\frac{10.{2}^{x}-1}{9+{({2}^{x})}^{2}+6.{2}^{x}}]={\mathrm{log}}_{2}(\frac{1}{2})$⇒log2​[9+(2x)2+6.2x10.2x−1​]=log2​(21​)
$\Rightarrow \frac{10.{2}^{x}-1}{9+{({2}^{x})}^{2}+6.{2}^{x}}=\frac{1}{2}$⇒9+(2x)2+6.2x10.2x−1​=21​
$\Rightarrow 2(10.{2}^{x}-1)=9+{({2}^{x})}^{2}+6.{2}^{x}$⇒2(10.2x−1)=9+(2x)2+6.2x
$\Rightarrow {({2}^{x})}^{2}-14.{2}^{x}+11=0$⇒(2x)2−14.2x+11=0
Let ${2}^{x}=y$2x=y
$\Rightarrow {y}^{2}-14y+11=0$⇒y2−14y+11=0
Let roots are {y}_{1}={2}^{{x}_{1}}&{y}_{2}={2}^{{x}_{2}}
Product of roots, ${y}_{1}{y}_{2}={2}^{{x}_{1}}\times {2}^{{x}_{2}}={2}^{{x}_{1}+{x}_{2}}=11$y1​y2​=2x1​×2x2​=2x1​+x2​=11
$\Rightarrow \mathrm{log}{2}^{{x}_{1}+{x}_{2}}=\mathrm{log}11$⇒log2x1​+x2​=log11
$\Rightarrow ({x}_{1}+{x}_{2})\mathrm{log}2=\mathrm{log}11$⇒(x1​+x2​)log2=log11
$\Rightarrow {x}_{1}+{x}_{2}=\frac{\mathrm{log}11}{\mathrm{log}2}$⇒x1​+x2​=log2log11​
Hence, ${x}_{1}+{x}_{2}={\mathrm{log}}_{2}11$x1​+x2​=log2​11