<s>[INST] So many logs with different bases 
$ \large { 6 }^{ \log _{ 5 }{ x }  }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x }  }\log _{ 3 }{ \frac { x }{ 3 }  } ={ 6 }^{ \log _{ 5 }{ 5x }  }-{ 5 }^{ \log _{ 6 }{ x }  }$
  The sum of the solutions to the equation above can be expressed as $a^{b/c} + d$ find $abc+d$. 

I can't find a way out... But if $x = 1$ then the equation is true. I think $d = 1$. I guess first I've to simplify the equation. But can't find any way... The bases of the logs are also different. any hint will be helpful.... 
P.S: Problem Collected from Brilliant.org 
 [/INST] $$ \large { 6 }^{ \log _{ 5 }{ x }  }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x }  }\log _{ 3 }{ \frac { x }{ 3 }  } ={ 6 }^{ \log _{ 5 }{ 5x }  }-{ 5 }^{ \log _{ 6 }{ x }  }$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{\log _{6}{6}+\log_6x}(\log_{3}x-\log_3 3)={6}^{\log _{ 5 }{5}+\log_5x}-{5}^{\log _{6}{x}}$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{1+\log_6x}(\log_{3}x-1)={6}^{1+\log_5x}-{5}^{\log _{6}{x}}$$
$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$
$$\large 5\cdot{6}^{\log _{5}{x} }\log _{3}{x}-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$
Let $\log_5x=a; \log_3x=b; \log_6x=c.$
$$\large 5\cdot{6}^{a}\cdot b-5\cdot{5}^{c}\cdot b+5\cdot{5}^{c}=6\cdot{6}^{a}-{5}^{c}$$
$$5b(6^a-5^c)=6(6^a-5^c)$$
$$(6^a-5^c)(5b-6)=0$$
Then $6^a-5^c=0$ or $5b-6=0$
Case 1)$6^a-5^c=0 \Leftrightarrow 6^{\log_5x}=5^{\log_6x}$
$$6^{\log_5x}=5^{\log_6x}\Leftrightarrow x=1$$
Case 2) $5\log_3x=6$
$$\log_3x=\frac65 \Rightarrow x=\sqrt[5]{3^6}$$
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