JEE Main 2021 — Basic of Mathematics Question with Solution

Given, $x+1-2{\log }_2\left(3+2^x\right)+2{\log }_4\left(10-2^{-x}\right)=0$

$\Rightarrow x+1-2{\log }_2\left(3+2^x\right)+2{\log }_{2^2}\left(\frac{10.2^x-1}{2^x}\right)=0$

$\Rightarrow x+1-2{\log }_2\left(3+2^x\right)+{\log }_2\left(\frac{10.2^x-1}{2^x}\right)=0$

$\Rightarrow x+1-2{\log }_2\left(3+2^x\right)+{\log }_2\left(10.2^x-1\right)-{\log }_2{2}^x=0$

$\Rightarrow x+1-{\log }_2{\left(3+2^x\right)}^2+{\log }_2\left(10.2^x-1\right)-x{\log }_{2}2=0$

$\Rightarrow x+1-{\log }_2{\left(3+2^x\right)}^2+{\log }_2\left(10.2^x-1\right)-x=0$

$\Rightarrow x+1+{\log }_2\left[\frac{10.2^x-1}{{\left(3+2^x\right)}^2}\right]-x=0$

$\Rightarrow 1+{\log }_2\left[\frac{10.2^x-1}{{\left(3+2^x\right)}^2}\right]=0$

$\Rightarrow {\log }_2\left[\frac{10.2^x-1}{{\left(3+2^x\right)}^2}\right]=-1$

$\Rightarrow {\log }_2\left[\frac{10.2^x-1}{9+{\left(2^x\right)}^2+6.2^x}\right]={\log }_2\left(\frac{1}{2}\right)$

$\Rightarrow \frac{10.2^x-1}{9+{\left(2^x\right)}^2+6.2^x}=\frac{1}{2}$

$\Rightarrow 2\left(10.2^x-1\right)=9+{\left(2^x\right)}^2+6.2^x$

$\Rightarrow {\left(2^x\right)}^2-14.2^x+11=0$

Let $2^x=y$

$\Rightarrow y^2-14y+11=0$

Let roots are $y_1=2^{x_1} \& y_2=2^{x_2}$

Product of roots, $y_1{y}_2=2^{x_1}\times 2^{x_2}=2^{x_1+x_2}=11$

$\Rightarrow \log 2^{x_1+x_2}=\log 11$

$\Rightarrow \left(x_1+x_2\right)\log 2=\log 11$

$\Rightarrow x_1+x_2=\frac{\log 11}{\log 2}$

Hence, $x_1+x_2={\log }_{2}11$