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  • The sum of the roots of the equation, <img alt="x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0" src="https://entrancecorner.codecogs.com/gif.latex?x&plus;1-2%20%5Clog%20_%7B2%7D%5Cleft%283&plus;2%5E%7Bx%7D%5Cright%29&p

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 :
Option: 1 \log _{2} 14
Option: 2 \log _{2} 12
Option: 3 \log _{2} 13
Option: 4 \log _{2} 11

Answers (1)

best_answer

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

\Rightarrow log_{2}\left ( 3+2^{x} \right )^{2}-\log_{2}\left ( 10 -2^{-x}\right )=x+1\\

\Rightarrow \log_{2}\left ( \frac{\left ( 2^{x} \right )^{2}+6.2^{x}+9}{10-2^{-x}} \right )= x+1\\

Raise to the power of 2, take  2^{x}=t

\Rightarrow \frac{t^{2}+6t+9}{10-\frac{1}{t}}=2t\\

\Rightarrow t^{2}+6t+9=20t-2\\

\Rightarrow t^{2}-14t+11=0< ^{t_{1}=2^{x_{1}}}_{t_{2}=2^{x_{2}}}\\

     t_{1}\cdot t_{2}=2^{x_{1}+x_{2}}=11\\

\Rightarrow x_{1}x_{2}=\log_{2}11

Posted by

Kuldeep Maurya

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