Question

The sum of the roots of the equation, $x+1-2{\log }_2(3+2^x)+2{\log }_4(10-2^{-x})=0,$ is

• A. ${\log }_{2}14$
• B. ${\log }_{2}13$
• C. ${\log }_{2}11$
• D. ${\log }_{2}12$

Answer 1

The correct option is C ${\log }_{2}11$

$x+1-2{\log }_2(3+2^x)+2{\log }_4(10-2^{-x})=0$
$\Rightarrow x+1-2{\log }_2(3+2^x)+2\frac{{\log }_2(10-2^{-x})}{{\log }_{2}4}=0$
$\Rightarrow x+1+{\log }_2(10-2^{-x})-{\log }_2(3+2^x{)}^2=0$
$\Rightarrow x+1-{\log }_2\left(\frac{(3+2^x{)}^2}{10-2^{-x}}\right)=0$
$\Rightarrow 2^{x+1}=\frac{9+6\cdot 2^x+2^{2x}}{10-2^{-x}}$
$\Rightarrow 20\cdot 2^x-2=9+6\cdot 2^x+2^{2x}$
$\Rightarrow (2^x{)}^2-14(2^x)+11=0$
Let two roots be $2^{x_1}$ and $2^{x_2}$
Then product of the roots:
$2^{x_1}\cdot 2^{x_2}=11\Rightarrow 2^{x_1+x_2}=11\Rightarrow x_1+x_2={\log }_{2}11$
$\therefore$ Sum of roots $={\log }_{2}11$