2018 AMC 12A Problem #14

$log_{3x}4=log_{2x}8$

$\Rightarrow \frac{log_{2}4}{log_{2}(3x)}=\frac{log_{2}8}{log_{2}(2x)}$

$\Rightarrow log_{2}4\cdot log_{2}(2x)=log_{2}8\cdot log_{2}(3x)$

$\Rightarrow 2log_{2}(2x)=3log_{2}(3x)$

$\Rightarrow 2(log_{2}(2)+log_{2}(x))=3(log_{2}(3)+log_{2}(x))$

$\Rightarrow 2log_{2}(2)+2log_{2}(x)=3log_{2}(3)+3log_{2}(x)$

$\Rightarrow 2log_{2}(2)-3log_{2}(3)=log_{2}(x)$

$\Rightarrow log_{2}(2^2)-log_{2}(3^3)=log_{2}(x)$

$\Rightarrow log_{2}(\frac{2^2}{3^3})=log_{2}(x)$

$\Rightarrow 2^{log_{2}(\frac{2^2}{3^3})}=2^{log_{2}(x)}$

$\Rightarrow \frac{2^2}{3^3}=x$

$\Rightarrow \frac{4}{27}=x$

Thus the required answer is $27+4=31$ .