Question

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

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

Answer 1

Answer: (B)

$x+1-2{\log }_2\left(3+2^x\right)+2{\log }_4\left(10-2^{-x}\right)=0$
${\log }_2\left(2^{x+1}\right)-{\log }_2{\left(3+2^x\right)}^2+{\log }_2\left(10-2^{-x}\right)=0$
${\log }_2\left(\frac{2^{x+1}\cdot \left(10-2^{-x}\right)}{{\left(3+2^x\right)}^2}\right)=0$
$\frac{2\left(10.2^x-1\right)}{{\left(3+2^x\right)}^2}=1$
$\Rightarrow 20.2^x-2=9+2^{2x}+6.2^x$
$\therefore {\left(2^x\right)}^2-14\left(2^x\right)+11=0$
Roots are $2^{x_1}\&2^{x_2}$
$\therefore 2^{x_1}\cdot 2^{x_2}=11$
$x_1+x_2={\log }_2(11)$