Question

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

Answer 1

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

${\log }_{2}x+{\log }_2{(x+2)}^{1/2}=2$

$\therefore {\log }_{2}x\sqrt{x+2}=2$

$x\sqrt{x+2}=4$

$x^3+2x^2-16=0$ (By squaring on both sides)

$x^3-2x^2+4x^2-16=0$

$x^2(x-2)+4(x-2)(x+2)=0$

$(x-2)(x^2+4x+8)=0$

No roots for $x^2+4x+8=0$

$\therefore x=2$