Can this log question be simplified? $ { 2^{log_3 5}} -  {5^{log_3 2}}.$
I don't know any formula that can apply to it or is there a formula?? 
Even a hint will be helpful.

$$ { 2^{\log_3 5}} -  {5^{\log_3 2}}= \\ 3^{\log_3 { 2^{\log_3 5}} }-3^{\log_3 {5^{\log_3 2}}}= \\ 3^{\log_3 5 \cdot \log_3 { 2} }-3^{\log_3 2 \cdot \log_3 {5}}= \\ \left ( 3^{\log_3 5 }\right )^{\log_3 { 2}} -\left ( 3^{\log_3 2 }\right )^{\log_3 {5} }= \\ 5^{\log_3 2 }-2^{\log_2 5} \Rightarrow \\  \\ { 2^{\log_3 5}} -  {5^{\log_3 2}}=5^{\log_3 2 }-2^{\log_2 5} \Rightarrow  \\ 2 \cdot 2^{\log_3 5}=2 \cdot 5^{\log_3 2} \Rightarrow \\  \\ \\ 2^{\log_3 5}= 5^{\log_3 2}$$
$$$$
So we conclude that $${ 2^{\log_3 5}} -  {5^{\log_3 2}}= 5^{\log_3 2} -  {5^{\log_3 2}}=0$$ 
