$\begin{align}  & {{\log }_{4}}(x+2){{\log }_{x}}2=1 \\ & {{\log }_{{{2}^{2}}}}(x+2)\frac{1}{{{\log }_{2}}x}=1 \\ & \frac{1}{2}{{\log }_{2}}(x+2)\frac{1}{{{\log }_{2}}x}=1\text{    }\left| note:{{\log }_{{{a}^{n}}}}b=\tfrac{1}{n}{{\log }_{a}}b \right. \\ & \text{lo}{{\text{g}}_{2}}(x+2)=2{{\log }_{2}}x \\ & lo{{g}_{2}}(x+2)={{\log }_{2}}{{x}^{2}} \\ & \text{ }x+2={{x}^{2}} \\ & {{x}^{2}}-x-2=0 \\ & (x-2)(x+1)=0 \\ & x=2\text{ or }x=1 \\\end{align}$