Calculate: log_{10}(x)+log_{10}(x-1)=log_{10}(3x+12)

Expression: $\log_{10}(x)+\log_{10}(x-1)=\log_{10}(3x+12)$

Apply log rules

$x(x-1)=3x+12$

Solve $ x(x-1)=3x+12:{\quad}x=6,x=-2$

$x=6,x=-2$

The solution is

$x=6$