Question

The equation ${\log }_2(3-x)+{\log }_2(1-x)=3$ has

• A. One root
• B. Two roots
• C. Infinite roots
• D. No root

Answer 1

The correct option is A One root

${\log }_2(3-x)+{\log }_2(1-x)=3$ is defined for $3-x>0$ and $1-x>0$
$\Rightarrow x<3$ and $x<1$
$\therefore$ It is defined for $x<1$
${\log }_2(3-x)+{\log }_2(1-x)={\log }_2{2}^3$
$={\log }_{2}8$
${\log }_2(3-x)(1-x)={\log }_{2}8$
Since, $\log a+\log b=\log ab$
$(3-x)(1-x)=8$
$3-4x+x^2=8$
$x^2-4x-5=0$
$x^2-5x+x-5=0$
$x=5,-1$
But, $x<1\Rightarrow x=-1$
Only one root .