Question

If, ${\log }_{2}x=\frac{1}{2}{\log }_2(x+2),$ find the value of x.

Answer 1

${\log }_{2}x=\frac{1}{2}{\log }_2(x+2),$

$\Rightarrow 2{\log }_{2}x={\log }_2(x+2)$

$\Rightarrow {\log }_2{x}^2={\log }_2(x+2)$

$\Rightarrow x^2=x+2$

$\Rightarrow x^2-x-2=0$

Hence, $x=-1,2$

But for, $x=-1$;

$LHS={\log }_2-1$ which is not possible.

Hence $x=2$ is required value of $x$.