If $2{{\log }_{y}}x+2{{\log }_{x}}y=5$ show that ${{\log }_{y}}x$ is either ½ or 2. Hence find the all pair of values of x and y which satisfy above and the equation xy =27
$\begin{align} & 2{{\log }_{y}}x+2{{\log }_{x}}y=5 \\ & 2{{\log }_{y}}x+2\left( \frac{1}{{{\log }_{y}}x} \right)=5 \\ & 2{{({{\log }_{y}}x)}^{2}}+2=5{{\log }_{y}}x \\ & 2{{({{\log }_{y}}x)}^{2}}-5{{\log }_{y}}x+2=0 \\ & 2{{({{\log }_{y}}x)}^{2}}-4{{\log }_{y}}x-{{\log }_{y}}x+2=0 \\ & 2{{\log }_{y}}x(lo{{g}_{y}}x-2)-1({{\log }_{y}}x-2)=0 \\ & (2{{\log }_{y}}x-1)(lo{{g}_{y}}x-2)=0 \\ & {{\log }_{y}}x=\frac{1}{2}\text{ or }{{\log }_{y}}x=2 \\ & \text{if also }xy=27 \\ & When\text{ lo}{{\text{g}}_{y}}x=\tfrac{1}{2} \\ & {{\log }_{y}}x=\frac{1}{2} \\ & x={{y}^{\tfrac{1}{2}}} \\ & \text{Substitute }x={{y}^{\tfrac{1}{2}}}\text{ into }xy=27 \\ & {{y}^{\tfrac{1}{2}}}y=27 \\ & {{y}^{\tfrac{3}{2}}}=27 \\ & y={{27}^{\tfrac{2}{3}}}={{({{3}^{3}})}^{\tfrac{2}{3}}}={{3}^{2}}=9 \\ & x={{y}^{\tfrac{1}{2}}}={{9}^{\tfrac{1}{2}}}=\pm 3 \\ & (x,y)=(\pm 3,9) \\ & When\text{ lo}{{\text{g}}_{y}}x=2 \\ & x={{y}^{2}} \\ & \text{Substitute }x={{y}^{2}}\text{ into }xy=27 \\ & xy={{y}^{2}}y=27 \\ & {{y}^{3}}=27 \\ & y=3 \\ & x={{3}^{2}}=9 \\ & (x,y)=(9,3) \\ & \text{The dual value of }x\text{ and }y\text{ are }(x,y)=(\pm 3,9)\text{ and }(x,y)=(9,3) \\\end{align}$