Sum of Roots

The given equation reduces to the form :

$(\ln x) ^3=(\ln x) ^2(\ln 2+\ln 4+\ln 6)$

$\implies (\ln x) ^2(\ln x -\ln 48)=0$

$\implies x_1=1,x_2=48,x_1+x_2=\boxed {49}$ .

$\begin{aligned} \log_2 x \log_4 \log_6 x & = \log_2 x \log_4 x + \log_4 x \log_6 x + \log_6 x \log_2 x \\ \frac {\log^3 x}{\log 2 \log 4 \log 6} & = \frac {\log^2 x}{\log 2 \log 4} + \frac {\log^2 x}{\log 4 \log 6} + \frac {\log^2 x}{\log 6 \log 2} \\ \frac {\log^3 x}{\log 2 \log 4 \log 6} & = \frac {\log^2 x (\log 2 + \log 4 + \log 6)}{\log 2 \log 4 \log 6} \\ \log^3 x & = \log^2 x \log 48 \end{aligned}$

$\implies \log^2 x (\log x - \log 24) = 0 \implies \begin{cases} \log x = 0 & \implies x = 1 \\ \log x - \log 48 = 0 & \implies x = 48 \end{cases}$ .

Therefore the sum of roots is $1+48 = \boxed {49}$ .