$\begin{align}  & {{\log }_{x}}3+{{\log }_{3}}x=2.5 \\ & \text{lo}{{\text{g}}_{x}}3+\frac{1}{{{\log }_{x}}3}=\frac{5}{2}\left| {{\log }_{a}}b=\frac{1}{{{\log }_{b}}a}\text{Changing the base of a log} \right. \\ & \text{Let }{{\log }_{x}}3\text{ be }p \\ & p+\frac{1}{p}=\frac{5}{2} \\ & {{p}^{2}}+1=\frac{5p}{2} \\ & 2{{p}^{2}}-5p+2=0 \\ & 2{{p}^{2}}-4p-p+2=0 \\ & 2p(p-2)-1(p-2)=0 \\ & (2p-1)(p-2)=0 \\ & p=\frac{1}{2}\text{ or }p=2 \\ & \text{When }p=\frac{1}{2} \\ & {{\log }_{x}}3=\tfrac{1}{2} \\ & {{x}^{\tfrac{1}{2}}}=3 \\ & x={{3}^{2}}=9 \\ & \text{when }p=2 \\ & {{\log }_{x}}3=2 \\ & {{x}^{2}}=3 \\ & x=\sqrt{3} \\\end{align}$