Question

The sum of the roots of the equation. $x+1=2{log}_2(2^x+3)-2{log}_4(1980-2^{-x})is{log}_n(2^n+7)$.Find n

Answer 1

$x+1=2{\log }_2(2^x+3)-2{\log }_4(1980-2^{-x})$
$x+1=2{\log }_2(2^x+3)-{\log }_2(1980-2^{-x})$
$\Rightarrow x+1={\log }_2\frac{{(2^x+3)}^2}{(1980-2^{-x})}$
$2^{x+1}=\frac{{(2^x+3)}^2}{(1980-2^{-x})}$
$\Rightarrow 2^{2x}-{39542}^x+11=0$
Substitute $2^x=t$
$\Rightarrow t^2-3954t+11=0$
Let $t_1,t_2$ be the roots of the eqn.
$t_1{t}_2=11$
$\Rightarrow 2^{x_1}{2}^{x_2}=11$
$\Rightarrow 2^{x_1+x_2}=11$
$\Rightarrow x_1+x_2={\log }_{2}11$
But given sum of roots is ${\log }_n(2^n+7)$
So, on comparing, we get $n=2$