Logarithms

$\begin{aligned} \log_2 x \log_3 x \log_5 x & = \log_2 x \log_3 x +\log_2 x \log_5 x + \log_3 x \log_5 x \\ \frac {(\log x)^3}{\log 2 \log 3 \log 5} & = \frac {(\log x)^2}{\log 2 \log 3} + \frac {(\log x)^2}{\log 2 \log 5} + \frac {(\log x)^2}{\log 3 \log 5} \\ (\log x)^3 & = (\log 2 + \log 3 + \log 5)(\log x)^2 \\ & = \log 30 (\log x)^2 \\ (\log x)^2 (\log x - \log 30) & = 0 \end{aligned}$

$\implies \begin{cases} \log x = 0 & \implies x = 1 \\ \log x = \log 30 & \implies x = 30 \end{cases}$

Therefore, the sum of all the $x$ roots is $30+1=\boxed{31}$ .

When $x=1$ It's easy to see that both sides equal 0, hence 1 is a value of x that satisfy the equation When $x\ne1$ We can divide both sides by $\log _{ 2 }{ x } \log _{ 3 }{ x } \log _{ 5 }{ x }$ and we get $1=\frac { 1 }{ \log _{ 5 }{ x } } +\frac { 1 }{ \log _{ x }{ 3 } } +\frac { 1 }{ \log _{ x }{ 2 } }$ With the change of base formula we get $1=\log _{ x }{ 5 } +\log _{ x }{ 3 } +\log _{ x }{ 2 }$ Which can be expressed as $1=\log _{ x }{ (5\times 3\times 2) } =\log _{ x }{ 30 }$ Thus $x=30$ And the sum of both values of x is 31