The sum of the roots of the equation x+1-2 lo | vedclass

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(B) Given equation: $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$Using $\log _{4} a = \frac{1}{2} \log _{2} a$,we get:$x+1

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$\log _{2} 11$

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(B) Given equation: $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$Using $\log _{4} a = \frac{1}{2} \log _{2} a$,we get:$x+1-2 \log _{2}\left(3+2^{x}\right)+\log _{2}\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 (10-2^{-x})}{(3+2^{x})^{2}}\right)=0$$\frac{2 \cdot 2^{x} \cdot (10-2^{-x})}{(3+2^{x})^{2}}=1$$\frac{2(10 \cdot 2^{x}-1)}{(3+2^{x})^{2}}=1$$20 \cdot 2^{x}-2 = 9 + 2^{2x} + 6 \cdot 2^{x}$$(2^{x})^{2} - 14(2^{x}) + 11 = 0$Let $2^{x} = t$. Then $t^{2} - 14t + 11 = 0$. The roots are $t_{1} = 2^{x_{1}}$ and $t_{2} = 2^{x_{2}}$.From the product of roots of the quadratic equation,$t_{1} \cdot t_{2} = 11$.$2^{x_{1}} \cdot 2^{x_{2}} = 11 \Rightarrow 2^{x_{1}+x_{2}} = 11$$x_{1}+x_{2} = \log _{2}(11)$

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:

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