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

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Question

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

Vedclass · Accepted Answer

$\log _{2} 11$

Vedclass · Answer

$x+1-2 \log _{2}\left(3+2^{x}ight)+2 \log _{4}\left(10-2^{-x}ight)=0$

$\log _{2}\left(2^{x+1}ight)-\log _{2}\left(3+2^{x}ight)^{2}+\log _{2}\left(10-2^{-x}ight)=0$

$\log _{2}\left(\frac{2^{x+1} \cdot\left(10-2^{-x}ight)}{\left(3+2^{x}ight)^{2}}ight)=0$

$\frac{2\left(10.2^{x}-1ight)}{\left(3+2^{x}ight)^{2}}=1$

$\Rightarrow 20.2^{x}-2=9+2^{2 x}+6.2^{x}$

$	herefore\left(2^{x}ight)^{2}-14\left(2^{x}ight)+11=0$

Roots are $2^{x_{1}} \& 2^{x_{2}}$

$	herefore 2^{x_{1}} \cdot 2^{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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