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}}$
$= 2^{\log_3 5} - 5^{\frac{\log_5 2}{\log_5 3}}$ (change of base)
$= 2^{\log_3 5} - (5^{\log_5 2})^{(\frac{1}{\log_5 3})}$ (using $a^{\frac{b}{c}} = {(a^b)}^{\frac{1}{c}}$)
$= 2^{\log_3 5} - 2^{\frac{1}{\log_5 3}}$
$= 2^{\log_3 5} - 2^{\frac{\log_5 5}{\log_5 3}}$ ($\because \log_5 5 = 1$)
$=2^{\log_3 5} - 2^{\log_3 5} $ (basically a "reverse" change of base)
$=0$
