$\begin{align}  & {{\log }_{x}}4+{{\log }_{4}}x=2\tfrac{1}{2} \\ & \text{Note: }{{\log }_{a}}b=\frac{1}{{{\log }_{b}}a}\text{    }\!\!\{\!\!\text{ Change of base }\!\!\}\!\!\text{ } \\ & {{\log }_{x}}4+\frac{1}{{{\log }_{x}}4}=\frac{5}{2} \\ & \text{Let }{{\log }_{x}}4=a \\ & a+\frac{1}{a}=\frac{5}{2} \\ & 2{{a}^{2}}-5a+2=0 \\ & 2{{a}^{2}}-4a-a+2=0 \\ & 2a(a-2)-1(a-2)=0 \\ & (2a-1)(a-2)=0 \\ & a=\frac{1}{2}\text{ or }a=2 \\ & \text{When }a=\frac{1}{2} \\ & {{\log }_{x}}4=\frac{1}{2} \\ & {{x}^{\tfrac{1}{2}}}=4 \\ & x={{4}^{2}}=16 \\ & \text{when }a=2 \\ & {{\log }_{x}}4=2 \\ & {{x}^{2}}=4 \\ & x={{4}^{\tfrac{1}{2}}}=2 \\ & x=16\text{ or }x=2 \\\end{align}$