So many logs with different bases

Question

$ \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

Answer 1

$$ \large { 6 }^{ \log _{ 5 }{ x } }\log _{ 3 }( { x }^{ 5 } ) -{ 5 }^{ \log _{ 6 }{ 6x } }\log _{ 3 }{ \frac { x }{ 3 } } ={ 6 }^{ \log _{ 5 }{ 5x } }-{ 5 }^{ \log _{ 6 }{ x } }$$

$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{\log _{6}{6}+\log_6x}(\log_{3}x-\log_3 3)={6}^{\log _{ 5 }{5}+\log_5x}-{5}^{\log _{6}{x}}$$

$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-{5}^{1+\log_6x}(\log_{3}x-1)={6}^{1+\log_5x}-{5}^{\log _{6}{x}}$$

$$\large {6}^{\log _{5}{x} }\log _{3}({x}^{5})-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$

$$\large 5\cdot{6}^{\log _{5}{x} }\log _{3}{x}-5\cdot{5}^{\log_6x}\cdot\log_{3}x+5\cdot{5}^{\log_6x}=6\cdot{6}^{\log_5x}-{5}^{\log _{6}{x}}$$ 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) $5\log_3x=6$ $$\log_3x=\frac65 \Rightarrow x=\sqrt[5]{3^6}$$

Answer 2

$$\large 6^{\log_5 x} \log_3 (x^5) - 5^{\log_6 6x} \log_3 (x/3) = 6^{\log_5 5x} - 5^{\log_6 x}$$ $$\large 5 \cdot 6^{\log_5 x} \log_3 (x) - 5 \cdot 5^{\log_6 x} (\log_3 (x) - 1) = 6 \cdot6^{\log_5 x} - 5^{\log_6 x}$$ $$\large 6^{\log_5 x} (5 \log_3 (x) - 6) = 5^{\log_6 x} (5 \log_3 (x) - 6)$$

It is clear how to proceed now, so the work is left to you.