So Many Logs!

Relevant wiki: Logarithmic Functions - Problem Solving - Hard

Thus $abc+d=90+1=91$

Let $y=\log{(x)}$ where $log$ refers to the natural logarithm. The expression is equivalent to:

$\left(6^{\frac{y}{\log{ \left(y \right ) }}} \right ) \left( \frac{5y}{\log{ \left(3 \right ) }} \right ) - \left(5^{ \left(\frac{y}{\log{ \left(6 \right ) }}+1 \right ) } \right ) \left(\frac{y}{\log{ \left(3 \right ) }}-1 \right ) =6^{ \left(\frac{y}{\log{ \left(5 \right ) }}+1 \right ) }-5^{\frac{y}{\log{ \left(6 \right ) }}}$

$6^{\frac{y}{\log{ \left ( 5 \right ) }}} \left ( \frac{5y}{\log{ \left ( 3 \right ) }}-6 \right ) = 5^{\frac{y}{\log{ \left ( 6 \right ) }}} \left ( \frac{5y}{\log{ \left ( 3 \right ) }}-6 \right )$

$\left (6^{\frac{y}{\log{ \left ( 5 \right ) }}}-5^{\frac{y}{\log{ \left ( 6 \right ) }}} \right) \left ( \frac{5y}{\log{ \left ( 3 \right ) }}-6 \right ) =0$

$\left (6^{\frac{y}{\log{ \left ( 5 \right ) }}}-5^{\frac{y}{\log{ \left ( 6 \right ) }}} \right) =0 \: \mathit{ or } \: \left ( \frac{5y}{\log{ \left ( 3 \right ) }}-6 \right ) =0$

$\left ( \frac{5y}{\log{ \left ( 3 \right ) }}-6 \right ) =0 \Rightarrow y=\frac{6}{5} \log \left( 3 \right )=\log \left( 3^{\frac{6}{5}} \right )$

$6^{\frac{y}{\log{ \left ( 5 \right ) }}} - 5^{\frac{y}{\log{ \left ( 6 \right ) }}}=0 \Rightarrow 6^{\frac{y}{\log{ \left ( 5 \right ) }}} = 5^{\frac{y}{\log{ \left ( 6 \right ) }}} \Rightarrow \frac{y}{\log \left(5 \right )} \log \left(6 \right )=\frac{y}{\log \left(6 \right )} \log \left(5 \right )$

$\cdots \Rightarrow \log^2 \left (6 \right ) y=\log^2 \left (5 \right )y \Rightarrow y=0$

$y=0,y=\log \left( 3^{\frac{6}{5}} \right ) \Rightarrow x=1,x=3^{\frac{6}{5}}$

$3^{\frac{6}{5}}+1=a^{\frac{b}{c}}+d \Rightarrow a=3,b=6,c=5,d=1$

$abc+d=3 \times 6 \times 5 +1=\boxed{\boxed{91}}$

Note: I have used:

$\log_{a} \left (b \right)=\frac{\log \left(b \right )}{\log \left (a \right )}$