Question

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

Answer 1

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

${\log }_{2}x+{\log }_2{(x+2)}^{y_2}={\log }_2{2}^2$

${\log }_2\left[x\right(x+2{)}^{y_2}]={\log }_{2}4$

$x(x+2{)}^{y_2}=4$

$(x+2{)}^{y_2}=4/x$ [sequencing both sides]

$x+2=\frac{16}{x^2}$

$x^2(x+2)=16$

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

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

$x=2$