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.
 A: Use this:
$$x^{\log_by}=y^{\log_bx}.$$
A: A symmetrical approach:
$$(\log_n b)(\log_n a)=(\log_n a )(\log_n b)\\
\log_na^{log_n b}=\log_nb^{log_n a}\\
a^{log_n b}=b^{log_n a}$$
Put $a=2, b=5, n=3$:
$$2^{log_3 5}=5^{log_3 2}\\
\Rightarrow 2^{log_3 5}-5^{log_3 2}=0$$
A: Let $\displaystyle A=2^{\log_35}\implies \log A=\frac{\log 5}{\log 3}\log 2$
Similarly, $\displaystyle B=5^{\log_32}\implies\log B=\cdots$
$\displaystyle\implies\log B=\log A\implies\cdots$ as $A,B$ are real
A: $ { 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)
$=0$
A: $$ { 2^{\log_3 5}} -  {5^{\log_3 2}}= \\ 3^{\log_3 { 2^{\log_3 5}} }-3^{\log_3 {5^{\log_3 2}}}= \\ 3^{\log_3 5 \cdot \log_3 { 2} }-3^{\log_3 2 \cdot \log_3 {5}}= \\ \left ( 3^{\log_3 5 }\right )^{\log_3 { 2}} -\left ( 3^{\log_3 2 }\right )^{\log_3 {5} }= \\ 5^{\log_3 2 }-2^{\log_2 5} \Rightarrow \\  \\ { 2^{\log_3 5}} -  {5^{\log_3 2}}=5^{\log_3 2 }-2^{\log_2 5} \Rightarrow  \\ 2 \cdot 2^{\log_3 5}=2 \cdot 5^{\log_3 2} \Rightarrow \\  \\ \\ 2^{\log_3 5}= 5^{\log_3 2}$$
$$$$
So we conclude that $${ 2^{\log_3 5}} -  {5^{\log_3 2}}= 5^{\log_3 2} -  {5^{\log_3 2}}=0$$ 
