Question

Prove that the sum of the roots of the equation $x+1=2{\log }_2(2^x+3)-2{\log }_4(1980-2^{-x})$ is ${\log }_{2}11$

Answer 1

$x+1=2{\log }_2(2^x+3)-2{\log }_4(1980-2^{-x})$
$x+1={\log }_2(2^x+3{)}^2-2{\log }_{2^2}(1980-2^{-x})$ ($\because \log x^m=m\log x$)

$x+1={\log }_2(2^x+3{)}^2-{\log }_2(1980-2^{-x})$ ($\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$ )

$\Rightarrow x+1={\log }_2\frac{{(2^x+3)}^2}{(1980-2^{-x})}[\because \log a-\log b=\log \frac{a}{b}]$

Converting to exponential form,
$2^{x+1}=\frac{{(2^x+3)}^2}{(1980-2^{-x})}$

$\Rightarrow 2^{x+1}(1980-2^{-x})={(2^x+3)}^2$

$\Rightarrow 19802^{x+1}-2^{x+1-x}=2^{2x}+2\left(2^x\right)3+9$

$\Rightarrow 2^{2x}-39542^x+11=0$
Substitute $2^x=t$
$\Rightarrow t^2-3954t+11=0$
Let $t_1,t_2$ be the roots of the eqn.
Product of roots $t_1{t}_2=11$
$\Rightarrow 2^{x_1}{2}^{x_2}=11$
$\Rightarrow 2^{x_1+x_2}=11$
$\Rightarrow x_1+x_2={\log }_{2}11$