Question

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

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

Answer 1

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

$x+1-2{\log }_2(2^x+3)+2{\log }_4(10-2^{-x})=0$
$\Rightarrow x+1={\log }_2(2^x+3{)}^2-{\log }_2(10-2^{-x})$
$\Rightarrow x+1={\log }_2\frac{(2^x+3{)}^2}{(10-2^{-x})}$
$\Rightarrow 2\cdot 2^x=\frac{(2^x+3{)}^2}{(10-2^{-x})}$

Assuming $2^x=t$
$\Rightarrow 2t=\frac{(t+3{)}^2}{(10-t^{-1})}$
$\Rightarrow 20t-2=(t+3{)}^2$
$\Rightarrow t^2-14t+11=0\cdots \left(1\right)$

If $t_1=2^{x_1},t_2=2^{x_2}$ are roots of the equation $\left(1\right)$, then
$t_1{t}_2=2^{x_1}\cdot 2^{x_2}=11$
$\therefore x_1+x_2={\text{log}}_{2}11$