The sum of the roots of the equation

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 :

  1. $\log _{2} 14$

  2. $\log _{2} 11$

  3. $\log _{2} 12$

  4. $\log _{2} 13$


Correct Option: , 2

Solution:

$x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$

$\log _{2}\left(2^{x+1}\right)-\log _{2}\left(3+2^{x}\right)^{2}+\log _{2}\left(10-2^{-x}\right)=0$

$\log _{2}\left(\frac{2^{x+1} \cdot\left(10-2^{-x}\right)}{\left(3+2^{x}\right)^{2}}\right)=0$

$\frac{2\left(10.2^{x}-1\right)}{\left(3+2^{x}\right)^{2}}=1$

$\Rightarrow 20.2^{x}-2=9+2^{2 x}+6.2^{x}$

$\therefore\left(2^{x}\right)^{2}-14\left(2^{x}\right)+11=0$

Roots are $2^{\mathrm{x}_{1}} \& 2^{\mathrm{x}_{2}}$

$\therefore 2^{x_{1}} \cdot 2^{x_{2}}=11$

$x_{1}+x_{2}=\log _{2}(11)$

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